A ball is projected vertically upward with an initial velocity of 50 ms -1 at t =0 s. At t =2 s. another ball is projected vertically upward with same velocity. At t = s, second ball will meet the first ball ( g =10 ms -2).

Question

A ball is projected vertically upward with an initial velocity of 50 ms -1 at t =0 s. At t =2 s. another ball is projected vertically upward with same velocity. At t = s, second ball will meet the first ball ( g =10 ms -2). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let they meet at t = t So first ball gets t sec. 2 text nd gets ( t -2) sec . they will meet at same height h 1=50 t -(1/2) gt 2 h 2=50( t -2)-(1/2) g ( t -2)2 h 1= h 2 50 t -(1/2) gt 2=50( t -2)-(1/2) g ( t -2)2 100=(1/2) g [ t 2-( t -2)2] 100=(10/2)[4 t -4] 5= t -1 t =6 sec .

Q. A ball is projected vertically upward with an initial velocity of 50ms−1 at t=0s. At t=2s. another ball is projected vertically upward with same velocity. At t= ______s, second ball will meet the first ball (g=10ms−2).

Let they meet at t=t So first ball gets tsec. & 2nd gets (t−2)sec . & they will meet at same height h1​=50t−21​gt2 h2​=50(t−2)−21​g(t−2)2 h1​=h2​ 50t−21​gt2=50(t−2)−21​g(t−2)2 100=21​g[t2−(t−2)2] 100=210​[4t−4] 5=t−1 t=6sec.

Q. A ball is projected vertically upward with an initial velocity of $50 m s^{- 1}$ at $t = 0 s$ . At $t = 2 s$ . another ball is projected vertically upward with same velocity. At $t =$ ______ $s$ , second ball will meet the first ball $(g = 10 m s^{- 2})$ .