Question

♠ ANSWER THE ABOVE QUESTION..

Answer 1

Let the initial velocity of the ball be v.

In vertical direction:

Initial velocity in vertical direction (u) = 0 m/s

Height = 500 m

Acceleration due to gravity = 10 m/s^2

Let the time taken be t

using the second equation of motion:

$H = ut+\frac{1}{2}at^2$

$500 = 0 + \frac{1}{2} \times 10 \times t^2$

$t = \sqrt{\frac{500}{5} }$

t = 10 seconds

Now, the velocity of the ball in horizontal direction = $\frac{Distance}{time}$

$v = \frac{400}{10}$

v = 40 m/s

Now, using momentum conservation:

Initial momentum of ball and gun = 0 (at rest)

Let the velocity of the gun be V

Final momentum = 100 × V + 1 × 40 = Initial momentum = 0

100V = -40

$V = -\frac{40}{100}$

V = - 0.4 m/s

Hence, the velocity of the gun will be -0.4 m/s. The negative sign shows that it will be in the opposite direction.

Answer 2

Answer:

Same as the top answer...