Question

1. A gun with a muzzle velocity of 50m/s is fired straight into the air. Neglecting air resistance, what is the bullet’s maximum height (before beginning to fall)? (Assume that g, gravitational acceleration, is approximately 10m/s².)

a. 50m

b. 75m

c. none of these

d. 125m

e. 10m

2. Consider once again the bullet from the previous problem (1.). Which one of the following statements is true?

a. When the bullet is at its maximum height, its velocity will momentarily be zero but its acceleration will not be zero.

b. When the bullet is at its maximum height, its velocity and acceleration will both be nonzero.

c. When the bullet is at its maximum height, its velocity and acceleration will both momentarily be zero.

3. A gun with a muzzle velocity of 50m/s is fired horizontally (straight forward) from the top of a tower. Neglecting air resistance, how far below the tower’s top will the bullet be 3 seconds later? (Assume that g, gravitational acceleration, is approximately 10m/s².)

a. 600m

b. 300m

c. 150m

d. 10m

e. none of these

Answer 1

1 . Given ,

Initial velocity

u = 50m/s^2

At maximum height, V = 0 m/s^2

Using equation of kinematic.

V^2 = u^2+2gh

0^2 = 50^2+2(-10)h

h = 50^2/20 = 125 m .....Option (d) is correct.

b.}

Acceleration at at any known height never be zero. And at the height of 125 m...It is approx 10 m/s^2. Actually this is acceleration due to gravity.

Hence, When the bullet is at its maximum height, its velocity will momentarily be zero but its acceleration will not be zero.

C.) Given u = 50m/s

t = 3 s

Using equation of kinematics

V = u+gt

V = 50+10×3 = 80 m/s

Now, v^2 = u^2 +2gh

h = V^2-u^2/(2a)

h = (80^2-50^2)/(2*10)

h = 195 m Answer.