Answer :
Answer:
[tex]v_{ox}= 19.6\ m/s[/tex]
Explanation:
Data provided in the question:
Height above the ground, H= 5.0m
Range of the ball, R= 20 m
Initial horizontal velocity = [tex]v_{ox}[/tex]
Initial vertical velocity= [tex]v_{oy}[/tex] (Since ball was thrown horizontally only)
Acceleration acting horizontally, [tex]a_x[/tex] = 0 m/s² [ Since no acceleration acts horizontally) ]
Vertical Acceleration, [tex]a_y[/tex] = 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
[tex]H= v_{oy}t+\frac{1}{2}a_yt^2[/tex]
[tex]5= (0)t+\frac{1}{2}(9.8)t^2[/tex]
[tex]t= \frac{10}{9.8}=1.02 s[/tex]
Then using Equations of motion for horizontal motion,
[tex]R= v_{ox}t+\frac{1}{2}a_xt^2[/tex]
[tex]20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2[/tex]
[tex]v_{ox}= 19.6\ m/s[/tex]