A freight train is moving at a constant speed of 7 m/s. A man standing on a flat

Question

A freight train is moving at a constant speed of 7 m/s. A man standing on a flatcar throws a ball into the air and catches it as it falls. Relative to the flatcar, the initial velocity of the ball is 14.9 m/s straight up.

a.) What horizontal distance has the ball traveled by the time it is caught according to the man on the ground?

The answer isn't zero or 9.8 or -9.8..i already tried that and thats wrong..

Explanation / Answer

ans ) initial vertical velocity = 14.9 m/sec initial horizontal velocity = 7 m/sec height ball rises vertically v= u-gt [ v=0, u=14.9] => t = 1.52 sec [ time it takes to rise to hmax] hmax= u^2/2g = 11.32 m now it will fall back from same height s = ut+0.5gt^2 [u=0] =>t = 1.52 So net horizontal distance covered = 7(1.52+1.52)= 21.28 m

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