Answer:

Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity =
Initial vertical velocity=
(Since ball was thrown horizontally only)
Acceleration acting horizontally,
= 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration,
= 9.8 m/s² (Since only gravity acts on it)
Let 't' be the time taken to reach ground
Therefore, using equations of motion, we have



Then using Equations of motion for horizontal motion,


