Question

A bullet is fired from a gun at 30 angle to the horizontal with a muzzle velocity of 600 m/s.

Answer 1

Answer:

Part a)

Time to reach the maximum height is given as

$t = 30 s$

B)

Maximum height of the bullet is

$H = 4500 m$

C)

Horizontal range of the bullet is

$R = 31177 m$

Part d)

Final velocity of the bullet is v = 600 m/s at 30 degree

Explanation:

A) As we know that velocity of the bullet is

v = 600 m/s

$\theta = 30^0$

So we have

$v_y = v sin\theta$

$v_y = 600 sin30 = 300 m/s$

Now time to reach the maximum height is given as

$0 = 300 - g t$

$t = 30 s$

B) As we know that the object have zero final speed in y direction

$0 - v_y^2 = 2(-g) h$

$0 - 300^2 = (2)(-10)H$

$H = 4500 m$

C)

Now the horizontal displacement of the object in total time of flight is known as range of projectile

So we have

$R = v_x T$

$R = (600 cos30)(2 \times 30)$

$R = 31177 m$

Part d)

As we know that the final height is same as that of initial So it will strike the ground at same velocity as that of initial velocity

So we have

v = 600 m/s at 30 degree