A ball of mass is thrown at an angle of to the horizontal from the top of a 65 m high tower as shown in Fig. at . Another identical ball is thrown with velocity 20 m/s horizontally towards from the top of a 30 m high tower 1 s after the projection of the first ball. Both the balls move in the same vertical plane. If they collide in mid–air,

The two balls will collide at time |
|||||||

a) |
2 s |
b) |
5 s |
c) |
10 s |
d) |
3 s |

A ball of mass is thrown at an angle of to the horizontal from the top of a 65 m high tower as shown in Fig. at . Another identical ball is thrown with velocity 20 m/s horizontally towards from the top of a 30 m high tower 1 s after the projection of the first ball. Both the balls move in the same vertical plane. If they collide in mid–air,

The two balls will collide at time |
|||||||

a) |
2 s |
b) |
5 s |
c) |
10 s |
d) |
3 s |

1 Answer

127 votes

**(a)**

When two balls collide, their height from ground is same. Let it be . Then vertical displacement of the first and second balls is m and , respectively. Let the time of flight of the first ball up to the instant be seconds. Then the time of flight that that of second ball is seconds. Considering vertically downward component of motion of first ball,

(i)

For vertically downward component of motion of second ball,

(ii)

From Eqs.(i) and (ii),

Displacement Horizontal distance moved by first ball up to the instant of collision + that moved by the second ball up to the same instant

Just before collision, vertically downward component of velocity of first ball is

And that of second ball is

According to the law of conservation of momentum, vertically downward component of velocity of combined body is given by

or

Similarly, horizontally leftward component of velocity of combined body (just after collision) is given by

127 votes

127