A person skateboarding with a constant speed of 1.30 m/s releases a ball from a height of 1.25 m above the
ground. Given that x0 = 0 and y0 = h = 1.25 m, find x and y for (a) t = 0.250 s and (b) t = 0.500 s. (c) Find the velocity,
speed and direction of motion of the ball at t = 0.500 s.
Oh, skateboarding and physics, huh? Hang on, let me put on my science goggles and calculate some answers for you.
(a) At t = 0.250 s:
Since the person is skateboarding with a constant speed, the horizontal position (x) of the ball will be x = vxt = (1.30 m/s)(0.250 s). So x = 0.325 m.
For the vertical position (y), we can use the equation y = y0 + vy0t + 0.5gt^2. Since the ball was released from rest, vy0 = 0. Plugging in the values, we get y = 1.25 m + 0 - (0.5)(9.8 m/s^2)(0.250 s)^2. So y = 1.25 m - 0.30625 m. Therefore, y = 0.94375 m.
(b) At t = 0.500 s:
Using the same formula for horizontal position as before, x = (1.30 m/s)(0.500 s), therefore x = 0.65 m.
For the vertical position, y = 1.25 m + 0 - (0.5)(9.8 m/s^2)(0.500 s)^2. So y = 1.25 m - 0.6125 m, which gives us y = 0.6375 m.
(c) To find the velocity, we can use the formula vx = v0x and vy = v0y - gt, where v0x = 1.30 m/s and v0y = 0.
So vx = 1.30 m/s and vy = 0 - (9.8 m/s^2)(0.500 s). Therefore, vy = -4.9 m/s.
To find the speed, we can use the formula speed = √(vx^2 + vy^2). Plugging in the values, we get speed = √((1.30 m/s)^2 + (-4.9 m/s)^2). Ultimately, this gives us a speed of 5.04 m/s.
As for the direction of motion, since vy is negative, we can say that the ball is moving downward. Therefore, the direction of motion is downward.
And there you have it, the ball's position, velocity, speed, and direction at various times. Keep skateboarding and asking awesome questions!
To find the position of the ball at different times, we'll first find the horizontal position (x) and vertical position (y) separately.
We know that the horizontal velocity of the skateboarder is constant, so the horizontal position (x) can be found using the formula:
x = x0 + vt
where x0 is the initial horizontal position (which is 0 in this case), v is the horizontal velocity (1.30 m/s), and t is the time.
(a) For t = 0.250 s:
x = 0 + (1.30 m/s)(0.250 s)
x = 0.325 m
So the horizontal position at t = 0.250 s is 0.325 m.
To find the vertical position (y), we'll use the formula of linear motion under constant acceleration:
y = y0 + v0t + (1/2)at^2
where y0 is the initial vertical position (1.25 m), v0 is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
(a) For t = 0.250 s:
y = 1.25 m + (0 m/s)(0.250 s) + (1/2)(-9.8 m/s^2)(0.250 s)^2
y = 1.25 m
So the vertical position at t = 0.250 s is 1.25 m, which is the initial height.
(b) For t = 0.500 s:
x = 0 + (1.30 m/s)(0.500 s)
x = 0.650 m
So the horizontal position at t = 0.500 s is 0.650 m.
(b) For t = 0.500 s:
y = 1.25 m + (0 m/s)(0.500 s) + (1/2)(-9.8 m/s^2)(0.500 s)^2
y = 1.25 m + 0 m + (-4.9 m/s^2)(0.250 s^2)
y = 1.25 m + (-4.9 m/s^2)(0.0625 s^2)
y = 1.25 m - 0.1225 m
y = 1.1275 m
So the vertical position at t = 0.500 s is 1.1275 m.
(c) To find the velocity at t = 0.500 s, we'll use the formula:
v = v0 + at
where v0 is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
v = 0 + (-9.8 m/s^2)(0.500 s)
v = -4.9 m/s
The velocity at t = 0.500 s is -4.9 m/s. The negative sign indicates the direction of motion is downward.
To find the speed, we'll take the absolute value of the velocity:
speed = |v| = |-4.9 m/s| = 4.9 m/s
So the speed of the ball at t = 0.500 s is 4.9 m/s.
The direction of motion of the ball at t = 0.500 s is downward (negative velocity).
To solve this problem, we will first outline the steps involved in finding the position of the ball at a given time. We'll use the following kinematic equations of motion to calculate the position and velocity of the ball:
1. Position in the x-direction: x = x0 + vxt
2. Position in the y-direction: y = y0 + vyt - (1/2)gt^2
3. Velocity in the x-direction: vx = v0x
4. Velocity in the y-direction: vy = v0y - gt
5. Speed: speed = sqrt(vx^2 + vy^2)
6. Direction of motion: direction = atan(vy/vx)
Now let's use these equations to find the position, velocity, speed, and direction of the ball at different times.
(a) For t = 0.250 s:
Given:
x0 = 0
y0 = h = 1.25 m
v0x = 1.30 m/s
v0y = 0 m/s (since the ball is released vertically)
g = 9.8 m/s^2 (acceleration due to gravity)
Using equation 1, we can find the x-coordinate:
x = x0 + v0x * t
x = 0 + 1.30 * 0.250
x = 0.325 m
Using equation 2, we can find the y-coordinate:
y = y0 + v0y * t - (1/2) * g * t^2
y = 1.25 + 0 * 0.250 - (1/2) * 9.8 * (0.250)^2
y = 1.25 - 0.3035
y = 0.9465 m
(b) For t = 0.500 s:
Given:
x0 = 0
y0 = h = 1.25 m
v0x = 1.30 m/s
v0y = 0 m/s (since the ball is released vertically)
g = 9.8 m/s^2 (acceleration due to gravity)
Using equation 1, we can find the x-coordinate:
x = x0 + v0x * t
x = 0 + 1.30 * 0.500
x = 0.650 m
Using equation 2, we can find the y-coordinate:
y = y0 + v0y * t - (1/2) * g * t^2
y = 1.25 + 0 * 0.500 - (1/2) * 9.8 * (0.500)^2
y = 1.25 - 1.225
y = 0.025 m
(c) For t = 0.500 s:
Given:
v0x = 1.30 m/s
v0y = 0 m/s (since the ball is released vertically)
g = 9.8 m/s^2 (acceleration due to gravity)
Using equation 3, we can find the x-velocity:
vx = v0x
vx = 1.30 m/s
Using equation 4, we can find the y-velocity:
vy = v0y - gt
vy = 0 - 9.8 * 0.500
vy = -4.90 m/s
Using equations 5 and 6, we can find the speed and direction of motion:
speed = sqrt(vx^2 + vy^2)
speed = sqrt(1.30^2 + (-4.90)^2)
speed = sqrt(1.69 + 24.01)
speed = sqrt(25.70)
speed = 5.07 m/s
direction = atan(vy/vx)
direction = atan((-4.90)/1.30)
direction = atan(-3.77)
direction = -74.15 degrees (Note: direction is measured counterclockwise from the positive x-axis)
So, at t = 0.500 s, the ball's position is (x = 0.650 m, y = 0.025 m). Its velocity is (vx = 1.30 m/s, vy = -4.90 m/s), and its speed is 5.07 m/s. The direction of motion is -74.15 degrees with respect to the positive x-axis.
a. X = 1.30m/s * 0.25s = 0.325 m.
Y = yo - 0.5g*t^2.
Y = 1.25 - 4.9*(0.25)^2 = 0.944 m. above gnd.
b. X = 1.30m/s * 0.500s =
Y = 1.25 - 4.9*(0.50)^2 =
c. Vy = Vo + g*t.
Vy = 0 + 9.8*0.5 = 4.9 m/s. = Ver. component of velocity.
Vx = 1.30 m/s and remains constant.
V = Sqrt(Vx^2+Vy^2)
Tan A = Vy/Vx = 4.9/1.30 =
A = ?.