) A batter hits a baseball so that it leaves the bat at speed
Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2. (a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the Iwrizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.
Question:
A batter hits a baseball so that it leaves the bat at speed
Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2.
(a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s.
SOLUTION FOR A:
POSITION OF THE BALL:
In order to find the position of the ball, split up the Vo component into Vox and Voy. That gives you:
Vox = 37.0cos(57.1) =22.2 m/s
Voy = 37.0sin(57.1) = 29.6 m/s
Once you solve for those, your goal is to find the final positions of the x and y coordinates, and you use the kinematic equation for position when given time, acceleration, and velocity: x= xo+vt+1/2at^2 -->
Solving for X final:
Vox = 22.2 m/s
Xf=Xo+vt+1/2at^2, and intial position of x is 0 and since there is no acceleration in the X direction get rid of that last term (1/2at^2). So now you simplify to get the equation Xf=vt, which is just Xf= Vox *t --> Xf = 22.2(2)= 44.4m
Solving for y final:
Voy = 29.6 m/s
Yf=Yo+Vyo*t+1/2at^2, now again, Yo is 0 and so you simplify this equation down to Yf=Voy*t+1/2*at^2, and the reason you don't get rid of the last term for Y is because you have gravity, and gravity is given as g=9.8=a. Yf=29.6(2)+(1/2)(-9.8)(2^2). This is equal to 39.6.
Your final answer for position is (44.4, 39.6)
SOLVING FOR MAGNITUDE OF VELOCITY FOR THE BALL AT T=2:
Magnitudes are just the distances found by s=sqrt(x^2+y^2), and the magnitude of velocity is V(magnitude)=sqrt(Vxf^2+Vyf^2). We're solving for V(mag), you need to fine Vxf and Vyf.
Use the Vf=Vo+at:
for the Vxf:
Vxf=Vxo+at, cancel out at because once again, you don't have acceleration in the X component. You therefore get, Vxf=Vxo, and Vxo=22.2m/s, so Vxf=22.2m/s.
For the Vyf:
Vyf=Vyo+at, you have acceleration of gravity g=9.80m/s^2, and t=2, and plugging Vyo=29.6m/s, Vyf=29.6+(-9.8)(2), and that equals 10. Vyf=10m/s.
Vxf=22.2m/s.
Vyf=10m/s.
Now that you have Vxf and Vyf at t=2, you can solve for the magnitude of the velocity at t = 2 by V(magnitude)=sqrt(Vxf^2+Vyf^2) --> V(mag)= sqrt(22.2^2+10^2)=24.4 m/s
Magnitude of Velocity at time = 2 seconds is 24.4m/s
SOLVING FOR DIRECTION AT TIME 2 SECONDS:
The direction (or angle) is found by tan^-1(y/x), for the velocity it's therefore tan^-1(Vyf/Vxo)
Vxf=22.2m/s.
Vyf=10m/s.
Tan^-1(10/22.2)= 24.3º is your angle
The direction of velocity is 24.3º from the horizontal at t=2 secs.
ANSWER FOR A:
Position: (44.4, 39.6)
Magnitude of Velocity: 24.4m/s
Direction of Velocity: 24.3º
SOLVING FOR B:
(b) Find the time when the ball reaches the highest point of its flight and find its height h at this point.
To solve for the time when the ball has reached it's highest point, Vfy must equal 0, because if it's not longer moving upwards in the y direction that means it's reached it's max height. Use the second kinematics equation to find t given the initial and final velocities, and the acceleration due to gravity : Vyf=Vyo+at, substituting our knowns, a=-9.8, Vyf=0, and Vyo=29.6, we have 0=29.6+-9.8t, and therefore t=-29.6/-9.8, resulting in t=44.7 m. The time it takes for the ball to reach the highest point is 3.02 m.
t= 3.02 seconds
Solving for height, requires the first kinematics equation x=xo+vt+1/2at^2, and this turns into Yf=Yo+Vyo*t+(1/2)a*t^2, your knowns:
t=3.02
a=-98
Yo=0
Vyo=29.6
Your equation is
Yf=0+29.6(3.02)+(1/2)(-9.8)(3.02^2) =>
Yf=44.7 m
The height h at the highest point is 44.7 meters.
ANSWER TO B:
Time: 3.02 seconds
Height: 44.7 meters.
SOLVE FOR C:
(c) Find the horizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.
Since projectile motion in this problem will be symmetrical in distance, you can solve for the x distance traveled when it reaches it's maximum height at t=3.02 seconds using the first kinematic equation Xf=Xo+Voxt+1/2at^2, that is X=22.2(3.02), that gives you Xf=67.044, now multiply this by 2 to get the whole range, 67.044*2= 134
ANSWER TO C:
Range: 134 meters
Vo = 37 m/s
Theta = 53.1
Xo = 0 m i
Yo = 0 m j
ax = 0 m/s^2 i
ay = -9.8 m/s^2 j
Vox = Vo * costheta
= 37[cos(53.1)]
= +22.2 m/s i
Voy = Vo * sintheta
= 37[sin(53.1)]
Voy = +29.6 m/s j
x1 = +44.4 m i
(a) x = 37.0 cos53.1 * t = 22.22 t
y = 37.0 sin53.1 t - (g/2) t^2
= 29.59 t - 4.90 t^2
Vx = 22.22
Vy = 29.59 - 9.8 t
angle to horizontal = arctan Vy/Vx
Plug in t = 2 s
(b) Compute y at time when Vy = 0
t = Vyo/g = 3.02 s
y = (Vyo)^2/(2g)
(c) R = 2 Vo^2 sin53.1*cos53.1/g
= 0.96 Vo^2/g
Vo is the launch velocity and Vyo is the y component of the launch velocity
24.23m/s, 23.21 degree
(a) Well, it seems like this baseball is really going places! To find the position of the ball at t = 2.00 s, we'll need to break down its initial velocity into horizontal and vertical components.
The horizontal component Vx remains constant throughout the motion, so it's given by:
Vx = Vo * cos(ao)
Plugging in the values, we get:
Vx = 37.0 m/s * cos(53.1°)
Now, let's find the horizontal position:
x = Vx * t
Substituting the values, we have:
x = (37.0 m/s * cos(53.1°)) * 2.00 s
So, the position of the ball at t = 2.00 s is x = (37.0 m/s * cos(53.1°)) * 2.00 s. Crunch those numbers and you'll get your answer!
As for the magnitude and direction of the velocity, the vertical component Vy changes due to gravity. We can find Vy using this equation:
Vy = Vo * sin(ao) - g * t
Since the ball is at its highest point at some time, we know that Vy = 0 at that moment. We can use that to find the time it reaches the highest point.
(b) To find the time, we set Vy = 0:
0 = Vo * sin(ao) - g * t
Solve for t, and you'll find the time it takes for the ball to reach the highest point.
To find the height h at that point, we can use this equation:
h = Vo * sin(ao) * t - (1/2) * g * t^2
Plug in the values you found, and you'll get the height h!
(c) To find the horizontal range R, we can use the equation:
R = Vx * t
Plug in the values you know, and you'll have the horizontal range!
Hope that helps, and remember, baseballs may never reach the same heights as comedians, but they do go quite far!
To calculate the position of the ball, and the magnitude and direction of its velocity at t = 2.00 s, we can use the following kinematic equations:
1. Position of the ball:
The horizontal position can be calculated using the formula:
x = Vo * cos(ao) * t
The vertical position can be calculated using the formula:
y = Vo * sin(ao) * t - (1/2) * g * t^2
Substituting the given values, we have:
x = 37.0 m/s * cos(53.1°) * 2.00 s
y = 37.0 m/s * sin(53.1°) * 2.00 s - (1/2) * 9.80 m/s^2 * (2.00 s)^2
2. Magnitude and direction of velocity:
The magnitude of the velocity can be calculated using the formula:
v = sqrt(vx^2 + vy^2)
The direction of the velocity can be calculated using the formula:
θ = arctan(vy / vx)
Substituting the given values, we have:
vx = Vo * cos(ao)
vy = Vo * sin(ao)
Now, let's calculate these values step by step:
a) Position of the ball at t = 2.00 s:
x = 37.0 m/s * cos(53.1°) * 2.00 s
= 37.0 m/s * 0.604 * 2.00 s
= 44.84 m
y = 37.0 m/s * sin(53.1°) * 2.00 s - (1/2) * 9.80 m/s^2 * (2.00 s)^2
= 37.0 m/s * 0.794 * 2.00 s - (1/2) * 9.80 m/s^2 * 4.00 s^2
= 59.72 m - 39.2 m
= 20.52 m
Therefore, at t = 2.00 s, the position of the ball is x = 44.84 m and y = 20.52 m.
b) Time when the ball reaches the highest point of its flight and its height (h) at that point:
To find the time when the ball reaches the highest point, we can use the formula for vertical velocity:
vy = Vo * sin(ao) - g * t
At the highest point, vy = 0, so we can solve for t:
0 = Vo * sin(ao) - g * t
t = Vo * sin(ao) / g
Substituting the given values, we have:
t = 37.0 m/s * sin(53.1°) / 9.80 m/s^2
= 48.95 / 9.80
= 5.00 s
To find the height (h) at the highest point, we can use the formula for vertical position:
y = Vo * sin(ao) * t - (1/2) * g * t^2
Substituting the given values, we have:
h = 37.0 m/s * sin(53.1°) * 5.00 s - (1/2) * 9.80 m/s^2 * (5.00 s)^2
= 37.0 m/s * 0.794 * 5.00 s - (1/2) * 9.80 m/s^2 * 25.00 s^2
= 98.5 m - 306.25 m
= -207.75 m
Therefore, the ball reaches its highest point at t = 5.00 s and the height (h) at this point is -207.75 m.
c) Horizontal range (R):
The horizontal range is the distance traveled by the ball in the horizontal direction from the starting point to where it hits the ground. It can be calculated using the formula:
R = x * 2
Substituting the given values, we have:
R = 44.84 m * 2
= 89.68 m
Therefore, the horizontal range (R) is 89.68 m.
To solve this problem, we can break it down into three parts:
(a) Finding the position and velocity at t = 2.00 s
(b) Finding the time when the ball reaches the highest point and its height at that point
(c) Finding the horizontal range — the distance traveled by the ball from the starting point to where it hits the ground.
Let's tackle each part step by step:
(a) Finding the position and velocity at t = 2.00 s:
To find the position of the ball at a given time, we need to split the initial velocity into its horizontal and vertical components.
The horizontal component (Vox) remains constant throughout the motion. We can find Vox using the initial velocity and the given angle of projection:
Vox = Vo * cos(ao)
= 37.0 m/s * cos(53.1°)
The vertical component (Voy) changes due to acceleration due to gravity. We can find Voy using the initial velocity and the given angle of projection:
Voy = Vo * sin(ao)
= 37.0 m/s * sin(53.1°)
The position at t=2s can be calculated by:
x = Vox * t
y = Yo + Voy * t + 0.5 * g * t^2
Where Yo is the initial height, which is not given in the problem. If we assume Yo = 0, then:
x = Vox * t
y = Voy * t + 0.5 * g * t^2
Substituting the given values into the equations, we can find the position (x, y) and the magnitude and direction of the velocity:
x = Vox * t
y = Yo + Voy * t + 0.5 * g * t^2
(b) Finding the time when the ball reaches the highest point and its height at that point:
To find the time when the ball reaches the highest point, we need to find the time when the vertical component of velocity becomes zero.
Voy = 0 at the highest point, so we can use the equation:
Voy = Vo * sin(ao) - g * t
0 = 37.0 m/s * sin(53.1°) - 9.8 m/s^2 * t
Solving this equation for t will give us the time when the ball reaches the highest point.
To find the height at the highest point, we can use the equation for vertical displacement:
y = Yo + Voy * t + 0.5 * g * t^2
Substituting the values of t and solving for y will give us the height at the highest point.
(c) Finding the horizontal range:
The horizontal range R is the horizontal distance traveled by the ball.
Since the horizontal component of velocity remains constant throughout the motion, we can find the horizontal range using the equation:
R = Vox * t
Substituting the value of t will give us the horizontal range.
I hope this breakdown helps you in solving the problem step by step. Let me know if you have any questions or need further clarification!