A ball with mass kg is thrown upward with initial velocity m/s from the roof of a building m high. Neglect air resistance. Use m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. s

Explanation:

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A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 17 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground.

a) Using the equation of motion formula;

v² = u²+2gH where;

u is the initial velocity

v is the final velocity

theta is the angle of launch

g is the acceleration due to gravity.

H is the maximum height reached by the ball

Since the ball is thrown upwards, the acceleration due to gravity will be negative. The equation then becomes;

v² = u²-2gH

Given

v = 0m/s

u = 22m/s

g = 9.8m.s²

0² = 22²-2(9.8)H

-22² = -19.6H

H = -22²/-19.6

H = 24.69m

If the biuliding is 17m high, the maximum height above the ground that the ball reaches will be;

Hmax = 24.69+17

Hmax = 41.69m

b) The time it takes to hit the ground can be expressed using the formula

v = u-gt

0 = 22-9.8t

-22 = -9.8t

t = -22/9.8

t = 2.45secs

explanation:

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