1.

2 balls are thrown up. One ball is thrown 2 seconds after the second ball. If the initial velocity is 40m/s.Find the height at which they will meet

Answer»

Let FIRST ball is THROWN at t = 0s.
second ball is thrown at t = 2s
u = 40 m/s
G = -10 m/s²

let they meet after t seconds at a HEIGHT H. Then both the balls will be at H.
for 1st ball, H = UT+(1/2)gt²
for 2nd ball, H = u(t-2) + (1/2)g(t-2)²

ut+ \frac{1}{2}gt^2 =u(t-2)+ \frac{1}{2}g(t-2)^2 \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}g(t^2+4-4t) \\ \\ut+ \frac{1}{2}gt^2 =ut-2u+ \frac{1}{2}gt^2+2g-2gt \\ \\-2u+2g-2gt=0\\ \\u-g+gt=0\\ \\40-(-10)-10t=0\\ \\10t=50\\ \\t= \frac{50}{10}=5\ seconds

H = ut+ \frac{1}{2}gt^2 \\ \\H=40*5- \frac{1}{2}*10*5^2=\boxed{75\ m}


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