Question

use a(t)=-32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. How high will the ball go?

Answer 1

Answer:

144 ft

Step-by-step explanation:

Using a(t)=-32 feet per second squared as the acceleration due to gravity. A ball is thrown vertically upward from the ground with an initial velocity of 96 feet per second. Therefore, the ball can go 144 ft high.

Answer 2

Answer:

  144 ft

Step-by-step explanation:

The speed decreases linearly by 32 ft/sec/sec, so will be zero after ...

  (96 ft/s)/(32 ft/s²) = 3 s

The average speed over that period is half the initial speed, so the distance covered between the ground and the maximum height is ...

  (3 s)(96 ft/s)/2 = 144 ft

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Alternate solution

The relationship between velocity and distance for a constant acceleration is ...

  $v_f^{2}- v_i^{2} = 2ad$

For an initial velocity of 96 ft/s, a final velocity (at the top of travel) of 0, and the given acceleration, we have ...

  $d = \dfrac{v_f^{2}- v_i^{2}}{2a} = \dfrac{0-96^{2}}{2(-32)}= 144\,\dots\ \text{ft}$