Answer:
The height of the cliff is 27.02 m
Explanation:
Given;
height above the ground from which the arrow is shot, h₁ = 1.5 m
initial velocity of the projectile, v = 30 m/s
angle of projection, θ = 60⁰
time taken to reach top of the cliff, t = 4.0 s
The vertical component of the velocity is calculated as;
[tex]v_y = v \times sin(\theta)\\\\v_y = 30 \times sin(60)\\\\v_y = 25.98 \ m/s[/tex]
The height attained by the projectile at the given time is calculated as;
[tex]h_2= v_y t - \frac{1}{2}gt^2\\\\h_2 = 25.98\times 4 \ - \ \frac{1}{2} \times 9.8\times 4^2\\\\h_2= 103.92 \ - \ 78.4 \\\\h_2 = 25.52 \ m[/tex]
The height of the cliff is calculated as;
H = h₁ + h₂
H = 1.5 m + 25.52 m
H = 27.02 m
