A ball is thrown straight upward and returns to the thrower's hand after 2.20 s in the air. A second ball is thrown at an angle of 31.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?
You need the vertical velocities the same.
time in air for the second ball.
vertical velocity at the top is zero.
Vf=Vsin31-g*t
Vf=0, so t=Vsin31/g but this time is 1.1 seconds, so solve for V to make that happen.
26.7081046958
To determine the speed at which the second ball must be thrown, we can use the concept of projectile motion.
Let's assume that the vertical distance reached by both balls is 'h'.
For the first ball, which is thrown vertically, we can find the initial velocity (v) using the equation:
h = (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight (2.20 s in this case).
Now, for the second ball thrown at an angle of 31.0°, we need to find the initial velocity (V) that will give it the same vertical distance h.
The vertical motion equation for the second ball can be written as:
h = V * sin(31.0°) * t - (1/2) * g * t^2
Since h is the same for both balls, we can equate the two equations and solve for V:
(v) * (2.20 s) = (V) * sin(31.0°) * (2.20 s) - (1/2) * (9.8 m/s^2) * (2.20 s)^2
Simplifying the equation, we get:
v = V * sin(31.0°) - (9.8 m/s^2) * 2.20 s
Now we can solve for V:
V = (v + (9.8 m/s^2) * 2.20 s) / sin(31.0°)
Substituting the known values:
V = (0 m/s + (9.8 m/s^2) * 2.20 s) / sin(31.0°)
Calculating the value, we have:
V ≈ 20.8 m/s
Therefore, the second ball must be thrown at a speed of approximately 20.8 m/s in order to reach the same height as the first ball thrown vertically.
To solve this problem, we need to consider the motion of both balls and set up two equations: one for the vertical motion and one for the horizontal motion.
Let's start with the vertical motion equation for the ball thrown vertically (ball 1). Using the equation for motion under constant acceleration, we have:
y1 = v1*t1 + (1/2)*a1*t1^2
Where:
y1 = displacement of ball 1 (which is zero, since it returns to the thrower's hand)
v1 = initial velocity of ball 1
t1 = time taken by ball 1 to reach its maximum height (which is half of the total time in the air, so t1 = 2.20 s / 2 = 1.10 s)
a1 = acceleration due to gravity (which is -9.8 m/s^2, taking downward as negative)
Since the ball reaches the same height as ball 1, the maximum height reached by the second ball (ball 2) would also be zero. Thus, we can write the vertical motion equation for ball 2 as:
y2 = v2*t2 + (1/2)*a1*t2^2
Where:
y2 = 0 (maximum height reached by ball 2)
v2 = initial velocity of ball 2
t2 = total time taken by ball 2 to reach the same height (this is what we need to find)
Now, let's focus on the horizontal motion equation for ball 2. Since the horizontal motion is not affected by gravity, we can write:
x = v2*cos(theta)*t
Where:
x = horizontal displacement of ball 2 (which is also zero, since it returns to the original position)
v2 = initial velocity of ball 2
theta = angle of projection (given as 31.0°)
t = total time in the air for ball 2 (which is what we need to find)
Since the total time in the air for ball 2 is the same as the time taken to reach the maximum height (t1), we can equate the two equations for y2:
v2*t2 + (1/2)*a1*t2^2 = v1*t1 + (1/2)*a1*t1^2
Simplifying and substituting the known values:
(1/2)*(-9.8)*(t2^2) = 0 + (1/2)*(-9.8)*(1.10^2)
After canceling out the common factors and rearranging the equation, we get:
t2^2 = 1.10^2
t2 = 1.10 s
Now, let's solve the horizontal motion equation for ball 2 to find the initial velocity (v2):
0 = v2*cos(31.0°)*1.10
Simplifying and rearranging, we have:
v2*cos(31.0°) = 0
v2 = 0 / cos(31.0°)
v2 = 0
Therefore, the initial velocity of ball 2 must be zero in order to reach the same height as the ball thrown vertically.