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1st Equation of Motion

A ball is thr...

Question

A ball is thrown vertically upward from the ground with a speed of 24.5 m/s. After what time intervals, the ball will be at a height of 29.4m from the ground.

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Solution

Dear Student,
$A c c o r d i n g t o q u e s t i o n b y s e c o n d e q u a r i o n o f m o t i o n 29.4 = 24.5 t - \frac{1}{2} \times 9.8 \times t^{2} t^{2} - 5 t + 6 = 0 (t - 6) (t - 1) = 0 s o t_{1} = 1 s a n d t_{2} = 6 s s o a f t e r t_{2} - t_{1} = 6 - 1 = 5 s b a l l w i l l b e a t s a m e h e i g h t 29.4 m$
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