Question

A projectile is given an initial velocity of 5 m/s at an angle 30degree below horizontal from the top of a building 25 m high. Find :

(i) the time after which it hits the ground.

(ii) the distance from the building where it strikes the ground. (Take g = 10 m/s2)

Answer 1

Given info : A projectile is given an initial velocity of 5 m/s at an angle 30° below horizontal from the top of a building 25 m high.

To find :

1. the time after which it hits the ground.
2. the distance from the building where it strikes the ground.

solution : time of flight, T = 2usinФ/g

here initial velocity of the projectile = u = 5 m/s

angle of projection = Ф = 30°

acceleration due to gravity = g = 10 m/s²

so, T = (2 × 5 × sin30°)/10 = 0.5 sec

therefore the time after which it hits the ground is 0.5 sec.

the distance from the building where it strikes the ground = horizontal range = u²sin2Ф/g

= (5² × sin60°)/10 = (25 × √3/2)/10 = 2.165 m

Answer 2

Given:

A projectile is given an initial velocity of 5 m/s at an angle 30degree below horizontal from the top of a building 25 m high.

To find:

• Time after which it hits ground?
• Distance from building where it hits ground?

Calculation:

First, find the components of initial velocity:

Velocity in X axis:

u_(x) = u × cos(30°) = 5 × (√3)/2 = (5√3)/2 m/s.

Velocity in Y axis:

u_(y) = u × sin(30°) = 5 × 1/2 = 5/2 m/s.

Now, let time to reach ground be t :

$h = u_{y}(t) + \dfrac{1}{2} g {t}^{2}$

$\implies 25 = \dfrac{5}{2} (t) + 5 {t}^{2}$

$\implies 50 = 5 (t) + 10 {t}^{2}$

$\implies 10 = t + 2 {t}^{2}$

$\implies 2 {t}^{2} + t - 10 = 0$

$\implies 2 {t}^{2} + 5t - 4t- 10 = 0$

$\implies t(2t+ 5 ) - 2(2t + 5)= 0$

$\implies (t - 2)(2t + 5)= 0$

$\implies \: t - 2 = 0$

$\implies \: t = 2sec$

So, time taken is 2 sec

Now, distance from building is :

$d = u_{x}(t)$

$\implies d = \dfrac{5 \sqrt{3} }{2} \times 2$

$\implies d = \dfrac{10\sqrt{3} }{2}$

$\implies d = 5 \sqrt{3} \: metres$

So, it hits 5√3 metres on ground from base of building.