Question

A ball is launched at 30 meters per second (m/s) from a 20-meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = -4.9t^2 + 30t + 20, where S is in meters.
a. What is the maximum height the ball reaches?


b. At what time does the ball reach its maximum height?


c. At what time does the ball hit the ground?

Answer 1

The maximum height the ball reaches is 65.92 m.

The time taken for the ball to reach the maximum height is 3.06 seconds.

The time taken for the ball to hit the ground is 6.12 seconds.

maximum height the ball reaches

The maximum height the ball reaches is calculated as follows;

s(t) = -4.9t² + 30t + 20

at maximum height, final velocity = 0

v = 0

ds/dt = -9.8t + 30 = 0

9.8t = 30

t = 30/9.8

t = 3.06 seconds

s(3.06) = -4.9(3.06)²  + 30(3.06)  + 20

s(3.06) = 65.92 m

Thus, the time taken for the ball to reach the maximum height is 3.06 seconds.

Time of flight of the ball

The time taken for the ball to hit is the total time of flight = 2 x 3.06 seconds = 6.12 seconds.

Thus, the maximum height the ball reaches is 65.92 m.

The time taken for the ball to reach the maximum height is 3.06 seconds.

The time taken for the ball to hit the ground is 6.12 seconds.

Learn more about maximum height here: brainly.com/question/12446886

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