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A rock is thrown into the air going 15 m/s. How high does it go? Use the following equation to solve for the answer. vf2 = vi2 + 2aΔy (use g=-10 instead of -9.8)

User Ricconnect
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1 Answer

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The highest distance covered by the rock when thrown at a speed of 15 m / s is 11.25 m.

Step-by-step explanation:

The study of objects in motion and their inter-relationships are known as Kinematics. There are four kinematic equations, which can relate to displacement D, velocity v, time t, and acceleration a.

Due to gravity, an object undergoing projectile motion is subject only to the acceleration. Because this acceleration is constant in the kinematic equations.

Here we apply the equation,

v^2 = u^2 + 2*a*s

where,

v represents the Final speed at the top most point (v) = 0

u represents the Initial speed (u) = 15 m/s

Use g = -10 = a (given)

v^2 = u^2 + 2*a*s

( or) distance s = (v^2 - u^2) /2 * a

s = (0 - 225)/(2 x -10)

Highest distance covered s = 225 / 20 = 11.25 m.

User Sarahjane
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