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You throw a ball straight up into the air at a velocity of 15 m/s. You want to know how high above your hand the ball will be at exactly 2.5 sec after you released it.

Of the 5 motion variables - Vi, Vf, a, ΔX, and T, identify which one you are solving for and the magnitudes and directions of all of the others that are known

Which of the motion equations is best to use to determine the height of the ball?

How high is the ball above your hand at 2.5 sec after you throw it?

Respuesta :

SaniShahbaz

Answer:

We need to find height of the ball above hand = ΔX = ?

Initial velocity Vi is given = 15 m/s in the upward direction

Gravitational acceleration = a = g = 9.81 m/s² downward direction

Time to reach to the height ΔX = T = 2.5 s

Vf is not given

Second equation of motion

         [tex]h = v_{i}t + \frac{1}{2}gt^{2}[/tex]

ΔX = 6.84 m

Explanation:

Given data:

Vi = 15 m/s

T = 2.5 s

ΔX = ?

Using second equation of motion

                       ΔX = ViT - 0.5gT²

                       ΔX = (15)(2.5) - (0.5)(9.81)(2.5)²

                       ΔX = 37.5 - 30.66

                      ΔX = 6.84 m

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