A ball is thrown vertically downward with speed 5m/s, another ball is thrown vertically down after 2s. The velocity of second ball so that it meets first ball after 2 s of its throw is ?
Since the second ball is thrown two seconds later than the first ball and has to meet it after two more seconds, so the time taken by the first ball = 4 s and that by second ball = 2 s.
Let the distance travelled by first ball be S1 and that by second ball = S2
A/c, S1=S2
Let the speed by which the second ball is thrown be u m/s.
Using the equation, S = ut + 1/2 * a* t2,
S1 = 5*4 + 1/2*10*16 = 100 m [ Since, the ball is thrown with the gravity, acceleration will be g = 10 m/s2 approximately]
S2 = u*2 + 1/2 * 10 * 4 = 2u + 20
So. 2u+20 = 100 m/s
Hence, speed of second ball = u = 40 m/s
Let the distance travelled by first ball be S1 and that by second ball = S2
A/c, S1=S2
Let the speed by which the second ball is thrown be u m/s.
Using the equation, S = ut + 1/2 * a* t2,
S1 = 5*4 + 1/2*10*16 = 100 m [ Since, the ball is thrown with the gravity, acceleration will be g = 10 m/s2 approximately]
S2 = u*2 + 1/2 * 10 * 4 = 2u + 20
So. 2u+20 = 100 m/s
Hence, speed of second ball = u = 40 m/s