Steven and Danny are standing apart from each other throwing a ball. Steven thro

Question

Steven and Danny are standing apart from each other throwing a ball. Steven throws the ball to Danny and Danny throws it back immediately. Steven throws the ball with a velocity of 5.00 m/s and after 2.9 s he got the ball back. Suppose there's no deceleration of the ball when it's moving and that we can neglect the time for Danny to throw the ball. The speed of the ball when it is thrown back to Steven was 3.30 m/s. How far are they apart from each other?

Explanation / Answer

Total time = 2.5 sec Distance traveled by the ball is 2D (Dis separation between them) => 2D = 4.9 * t1 + 3.3*t2 also D = 4.9t1 D = 3.3t2 => t1 = 49/33 t2 and t1+t2 = 2.5 => 49/33 t2 + t2 = 2.5 => t2 = 1s => t1 = 1.5 s => D = 4.9* 1 = 4.9 m

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