Question

a ball is thrown upward with a speed of 15m/s. How high will it go before it begins to fall?[g=9.8m/s²]
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Answer 1

Explanation:

let consider that the ball at the it's max hight= S

so,

initial velocity = 0 m/s

finel velocity = 15 m/s

acceleration = 9.8 m/s^2

now,

${v}^{2} = {u}^{2} + 2as \\ 15 = 0 + 2 \times 9.8 \times s \\ s = \frac{15}{(2 \times 9.8)} \\ s = 0.76$

so it cover =0.76 m in the hight above the ground.