Roger tosses a ball straight upward at v. Ignore air drag?
To calculate the ball's motion when it is tossed straight upward by Roger at a velocity v, we can use the laws of physics and kinematics. Let's break down the process step by step:
1. Define the variables:
- v: Initial velocity of the ball (upward)
- a: Acceleration due to gravity (a constant value of approximately 9.8 m/s^2 on Earth)
- t: Time (duration of the ball's motion)
- d: Displacement (vertical distance covered by the ball)
2. Determine the initial velocity:
Since Roger is tossing the ball straight upward, the initial velocity (v) is positive.
3. Calculate the time it takes for the ball to reach maximum height:
When the ball reaches its maximum height, its final velocity becomes zero (since it momentarily stops before falling back down). We can use the kinematic equation:
v = u + at
0 = v + (-9.8)t
This equation gives us the time (t) it takes for the initial upward velocity (v) to reduce to zero.
4. Determine the maximum height reached by the ball:
To find the maximum height, we can use the kinematic equation:
d = ut + (1/2)at^2
Here, d represents the maximum height obtained by the ball.
5. Calculate the time it takes for the ball to fall back down:
The time it takes for the ball to return to its initial position (when it was tossed) is twice the time calculated in step 3.
By following these steps, you can calculate the ball's motion when Roger tosses it straight upward with a velocity of v, assuming there is no air drag.
say Vi, v initial, not v
say initial height of Roger's hand is Hi
a = - g = -9.8 meters/second^2 or -32 ft/second^2 or whatever
v = Vi - g t
h = Hi + Vi t - (1/2) g t^2
at top when v = 0
t = Vi/g at top
comes back down to Hi at 2 t