Question

A man stands at the edge of a building and throws his cellphone forward over the edge in rage with a speed of 21.7 m/s. The phone leaves his hand at a height of H = 73.0 m above level ground at the bottom of the building. The origin at the bottom of the building directly below where the phone leaves the hand. Find the time it takes to hit the ground below and the velocity and direction of the impact.

Answer 1

Answer:

Explanation:

v(speed)=21.7m/s

h=73m

g or a=10m/s^2                   (can also be taken as 9.8m/s^2)

Assuming that the phone was thrown in the direction perpendicular to the ground as it is mentioned that he threw the phone in the forward direction,

u(initial velocity)=0m/s, in the downwards direction

Using the 3rd equation of motion,

v=2$\sqrt{365\\}$m/s, in the downwards direction (also the direction of impact)

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