Question

A projectile is fired over horizontal ground from the origin with an initial speed of 60 m/s. What firing angles will produce a range of 300 m?

Answer 1

Answer:

$54.75^\circ$

Step-by-step explanation:

Formula for horizontal distance or range is given as:

$s=\dfrac{2v_i^2sin\theta cos\theta}{g}$

Where

$v_i$ is the initial velocity

$\theta$ is the angle with horizontal at which the projectile was thrown

$g$ is the acceleration due to gravity.

Given,

Initial velocity, $v_i$ = 60 m/s

Range, $s$ = 300 m

To find: $\theta = ?$

Solution:

Using the formula as stated above:

$300=\dfrac{2\times 60^2sin\theta cos\theta}{9.8}\\\Rightarrow 2sin\theta cos\theta =\dfrac{300\times 9.8}{3600}\\\Rightarrow sin2\theta =0.81666\\\Rightarrow \theta \approx \bold{54.75^\circ}$

Firing angle $54.75^\circ$ will produce a range of 300 m.