Answer:
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.
<u>We are given:</u>
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]
<u>Solving for time taken:</u>
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
<u>Final Horizontal velocity of the ball</u>
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