A ball is thrown directly upward from ground level with an initial velocity v(0) = 96 ft=s.
Assuming that the ball’s only acceleration is that due to gravity (i.e., a(t) = ��32 ft=s2),
determine the maximum height reached by the ball and the time it takes to return to ground
level.
This is calculus?
height= vo*time-32*t^2
dheight/dt= vo-64t
t= 96/64 seconds.
max height then is the height(96/64). Calculate.
To determine the maximum height reached by the ball and the time it takes to return to ground level, we can use the equations of motion.
Let's break down the problem step by step:
Step 1: Determine the time it takes for the ball to reach its maximum height.
We'll use the equation of motion for displacement:
s(t) = s(0) + v(0)t + (1/2)a(t^2)
In this case, the initial displacement s(0) is 0 (as the ball is thrown from ground level), the initial velocity v(0) is 96 ft/s, and the acceleration a is -32 ft/s^2 (negative because it acts opposite to the direction of motion).
We are interested in finding the time it takes for the ball to reach its maximum height, so we set the final displacement s(t) to zero (at the maximum height):
0 = 0 + 96t - 16t^2
Simplifying the equation, we get:
16t^2 - 96t = 0
Factoring out common terms, we have:
16t(t - 6) = 0
Solving for t, we have two possible solutions: t = 0 (the starting time) and t = 6 seconds. Since we are looking for the time it takes for the ball to reach its maximum height, we discard the solution t = 0. Therefore, the time it takes for the ball to reach its maximum height is 6 seconds.
Step 2: Determine the maximum height reached by the ball.
To find the maximum height, we substitute the time t = 6 seconds into the equation of motion for displacement:
s(t) = s(0) + v(0)t + (1/2)a(t^2)
s(6) = 0 + 96(6) - 16(6^2)
s(6) = 0 + 576 - 576
s(6) = 0
Therefore, the maximum height reached by the ball is 0 feet. This implies that the ball reaches its maximum height and then falls back to the ground level.
Step 3: Determine the time it takes for the ball to return to ground level.
We'll use the equation of motion for displacement again:
s(t) = s(0) + v(0)t + (1/2)a(t^2)
In this case, the initial displacement s(0), initial velocity v(0), and acceleration a remain the same. However, we are interested in finding the time it takes for the displacement s(t) to become zero again (return to ground level).
0 = 0 + 96t - 16t^2
Simplifying the equation, we get:
16t^2 - 96t = 0
Factoring out common terms, we have:
16t(t - 6) = 0
Solving for t, we have two possible solutions: t = 0 (the starting time) and t = 6 seconds. Since the ball is returning to the ground level after reaching the maximum height, we discard the solution t = 0. Therefore, the time it takes for the ball to return to ground level is also 6 seconds.
In summary:
The maximum height reached by the ball is 0 feet.
The time it takes for the ball to reach its maximum height is 6 seconds.
The time it takes for the ball to return to ground level is also 6 seconds.