A circus wishes to develop a new clown act. Fig. (1) shows a diagram of the proposed setup. A clown will be shot out of a cannot with velocity `v_(0)` at a trajectory that makes an angle `theta=45^(@)` with the ground. At this angile, the clown will travell a maximum horizontal distance. The cannot will accelerate the clown by applying a constant force of `10, 000N` over a very short time of `0.24s`. The height above the ground at which the clown begins his trajectory is `10m`.

A large hoop is to be suspended from the celling by a massless cable at just the right place so that the clown will be able to dive through it when he reaches a maximum height above the ground. After passing through the hoop he will then continue on his trajectory until arriving at the safety net. Fig (2) shows a graph of the vertical component of the clown's velocity as a function of time between the cannon and the hoop. Since the velocity depends on the mass of the particular clown performing the act, the graph shows data for serveral different masses.

If the mass of a clown doubles, his initial kinetic energy, `mv_(0)^(2)//2`, will :-

Options

A. Remain the same
B. Be reduce to half
C. Double
D. Four times

Answer

B

Explanation

Initial kinetic enery`=1//2 mV_(0)^(2)`
If mass doubles, then we can sec from `(v_(y)-t)` curve then velocity becomes half of previous.
`:. 1/2xx2mxx(v_(0)/2)^(2)=(1//2mv_(0)^(2))/2` Hence [B]