Question

A dart is thrown from 1.50 m high at 10.0 m/s toward a target 1.73 m from the ground. At what angle was the dart thrown?

Answer 1

The angle of projection is 12.26⁰.

Step-by-step explanation:

Given;

initial position of the dart, h₀ = 1.50 m

height above the ground reached by the dart, h₁ = 1.73 m

maximum height reached by the dart, Hm = h₁ - h₀ = 1.73 m - 1.50 m= 0.23 m

velocity of the dart, u = 10 m/s

The maximum height reached by the projectile is calculated as;

`H_m = (u^2sin^2 \theta)/(2g)`

where;

θ is angle of projection

g is acceleration due to gravity = 9.8 m/s²

`H_m = (u^2sin^2 \theta)/(2g)\\\\sin^2 \theta = (H_m \ * \ 2g)/(u^2) \\\\sin^2 \theta = (0.23 \ * \ 2(9.8))/(10^2) \\\\sin ^2\theta =0.04508\\\\sin \theta = √(0.04508) \\\\sin \theta = 0.2123\\\\\theta = sin^(-1)(0.2123)\\\\\theta = 12.26^0`

Therefore, the angle of projection is 12.26⁰.