Question

37.) A ball is thrown nearly vertically upward from a point near the corner of a tall building. It just misses the edge on the way down, and passes a point 50.0 m below its starting point 5.0 s after it leaves the throwers hand. (a) What was the initial velocity of the ball? (b) How high did it rise above its starting point? (c) What were the magnitude and direction of its velocity at the highest point?

Answer 1

Answer:

a) Initial velocity of the ball=14.5 m/s

b)The ball rise to a height of 10.72 m/s

c)The velocity of the ball is zero at the highest point.

Explanation:

Taking point at which the ball is thrown to be origin. Distanced measured to be positive in upwards directions.

a)

Now using equation of motion

$y=ut-\dfrac{gt^2}{2}\\-50=u\times5-\dfrac{9.8\times 5^2}{2}\\u=14.5\ \rm m/s$

The initial velocity of the ball is $u=14.5\ \rm m/s$

b) Let h be the height that ball rise above the starting point given by

$h=\dfrac{u^2}{2g}\\h=\dfrac{14.5^2}{2\times9.8}\\h=10.72\ \rm m$

c)The ball will be at rest with zero velocity at the highest point