Question

A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s, acceleration due to gravity is g = 10 m/s2, what is the maximum horizontal range covered?  ​

Answer 1

Answer:

Maximum horizontal range covered = 54.13 m

Explanation:

The formula for horizontal range is R = v2 (sin 2θ)/g.      

v = 25

θ = 60°

g = 10

The vertical component of the velocity is 25 · (√3)/2 = 12.5√3 m/s

The horizontal component of the velocity is 25 · ½ = 12.5 m/s

At g = 10 m/s² it will take 12.5√3 m/s / 10 m/s² = 1.25√3 s to reach zero upward velocity and the same amount of time to fall back 2 · 1.25√3 s = 2.5√3 s

That is the same time that the projectile has to fly horizontally at 12.5 m/s

12.5 m/s · 2.5√3 s = 31.25√3 ≈ 54.13 m

Answer 2

Given:

A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s.

To find:

Horizontal Range ?

Calculation:

• Horizontal Range refers to the horizontal distance travelled by a oblique projectile when thrown at an angle.

The general expression of RANGE is :

$R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies R = \dfrac{ {25}^{2} \times \sin(2 \times {60}^{ \circ} ) }{10}$

$\implies R = \dfrac{625 \times \sin( {120}^{ \circ} ) }{10}$

$\implies R = \dfrac{625 \times 0.866 }{10}$

$\implies R = 54.12 \: m$

So, range of projectile is 54.12 metres.