NEED HELP DUE AT 11:59!! A ball is thrown horizontally from the top of a building 130 m high. The ball strikes the
ground 53 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.

Since the ball was thrown horizontally, there was no vertical component in that force. and hence, the initial vertical velocity of the ball is 0 m/s and the initial horizontal velocity is r.

We are given:

initial velocity  (u) = 0 m/s     [vertical]

final velocity (v) = v m/s  [vertical]

time taken to reach the ground (t) = t seconds

acceleration (a) = 10 m/s/s   [vertical , due to gravity]

height from the ground (h) = 130 m

displacement (s) = 53 m [horizontal]

Solving for time taken:

From the third equation of motion:

s = ut + 1/2 at²

130 = (0)(t) + 1/2 * (10) * t²

130 = 5t²

t² = 26

t = √26 seconds  or   5.1 seconds

Final Horizontal velocity of the ball

Since the horizontal velocity of the ball will remain constant:

the ball covered 53 m in 5.1 seconds [horizontally]

horizontal velocity of the ball = horizontal distance covered / time taken

Velocity of the ball = 53 / 5.1

Velocity of the ball = 10.4 m/s

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be