Answer:
25.2 m
Explanation:
The horizontal distance traveled s = v × t×cosθ and t = s / (v×cos θ)
The vertical distance traveled h = v × t ×sin θ - 1/2 × g × t^2
Substituting for t, h = s×tan θ- 1/2 × g × s^2 / (v cos θ)^2
Now, Solve for v^2 and get v^2 = g × s^2 ÷ [2 ×cos^2θ × (s×tan θ - h)]
And v^2 = 9.8×3600 / [2×0.535×(60×0.933 - 25] =1065 and v = 32.6 m/s
As a check from the first equation t = 60 / (32.6×0.731) = 2.52 sec
Horizontal distance traveled s = 32.6×cos 43°×2.52 =60 m
Height reached 2×g×h = (v×sin43°)^2
⇒and h = 25.2 m (using 2×g×h =v×y ^2)
Since maximum height is reached at the edge of the cliff the projectile
will not travel beyond the cliff
