Roger tosses a ball straight upward at a velocity of +32m/s, calculate the maximum height of the ball and the time to get to the maximum height.
To calculate the maximum height of the ball and the time it takes to reach that height, we can use the equations of motion for vertical motion.
First, we need to determine the initial velocity, final velocity, acceleration, and time at the maximum height.
Given:
Initial velocity (u) = +32 m/s (positive because the direction is upward)
Final velocity (v) = 0 m/s (at the highest point, the ball momentarily stops before falling back down)
Acceleration (a) = -9.8 m/s^2 (negative because gravity acts in the opposite direction to the initial velocity)
Time at the maximum height (t) = ?
Using the equation v = u + at, we can solve for time:
0 = 32 - 9.8t
Rearranging the equation, we have:
9.8t = 32
Dividing both sides by 9.8:
t = 32/9.8 ≈ 3.27 seconds
Therefore, the time it takes for the ball to reach the maximum height is approximately 3.27 seconds.
To calculate the maximum height, we can use the equation s = ut + 0.5at^2, where s is the displacement (height), u is the initial velocity, t is the time, and a is the acceleration:
s = (32)(3.27) + 0.5(-9.8)(3.27)^2
Simplifying the equation:
s = 104.64 - 51.09
s ≈ 53.55 meters
Hence, the maximum height of the ball is approximately 53.55 meters.
V = Vo + g*Tr = 0.
32 - 9.8Tr = 0
Tr = 3.27 s. = Rise time or time to reach max ht.
h = Vo*Tr + 0.5g*Tr^2.
Vo = 32 m/s, Tr = 3.27 s., g = -9.8 m/s^2, h = ?.