Question

A bullet is fired with speed 50 m/s at 45degree angle find the height of the bullet when its direction of motion makes angle 30degree with the horizontal (correct answer is 125/3)

Answer 1

50cos45° = vcos 30°
v = 50✓2/√3

$v { }^{2} = u {}^{2} + 2as$

2500*2/3*(sin 30°)^2 = 2500*( sin 45°)^2 - 2* g*s

( * denotes multiplication)

Answer 2

Answer: The height of the bullet is $h = \dfrac{125}{3}m$.

Explanation:

Given that,

Speed = 50 m/s

We know that,

The horizontal velocity of the particle is the same at every point.

So, the horizontal velocity

$50 cos 45^{0} = v cos 30^{0}$

$v = \dfrac{50\times1.414}{1.732}$

$v = 40.82 m/s$

Now, using equation of motion

$v^{2}=u^{2}-2gh$

The velocity at height h is

$v^{2}sin^{2}30=(50)^{2}sin^{2}45-2\times 10\times h$

$(40.82)^{2}sin^{2}30=(50)^{2}sin^{2}45-2\times 10\times h$

$416.57= 1250-20 h$

$h = 41.66 m$

$h = \dfrac{125}{3}m$

Hence, the height of the bullet is $h = \dfrac{125}{3}m$.