A bullet is fired from a gun at 30 angle to the horizontal with a muzzle velocity of 600 m/s.
A)Find the time to reach the maximum height.
B)Find the maximum height.
c)Find the range
d)Find the velocity just before re-striking the ground. (Ans: 600 m/s at 30 ο
below the horizontal)

Part a)

Time to reach the maximum height is given as

B)

Maximum height of the bullet is

C)

Horizontal range of the bullet is

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

So we have

Now time to reach the maximum height is given as

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

C)

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

So we have

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

your answer is, orbit around the nucleus

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over