Answer:
the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
Explanation:
Given;
height of the cliff, h = 210 m
initial horizontal velocity of the cannonball, Ux = 50 m/s
initial vertical velocity of the cannonball, Uy = 0
The time for the cannonball to reach the ground is calculated as;
[tex]h = u_yt - \frac{1}{2} gt^2\\\\h = 0 - \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 210}{9.8} }\\\\t = 6.55 \ s[/tex]
The horizontal distance covered by the cannonball before it hits the ground is calculated as;
[tex]X = U_x \times \ t\\\\X = 50 \times \ 6.55\\\\X = 327.5 \ m[/tex]
Therefore, the horizontal distance covered by the cannonball before it hits the ground is 327.5 m
