Two stones art thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with a time lapse of 3 second, height at which they collide is

Question

Two stones art thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with a time lapse of 3 second, height at which they collide is (A) 58.8 m (B) 111.5 m (C) 117.6 m (D) 122.5 m

Tardigrade Admin · Accepted Answer

s = ut - (1/2) gt2 = u(t - 3) - (1/2) g(t - 3)2

Q. Two stones art thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with a time lapse of 3 second, height at which they collide is

58.8 m

111.5 m

117.6 m

122.5 m

$s = u t - \frac{1}{2} g t^{2} = u (t - 3) - \frac{1}{2} g (t - 3)^{2}$

Q. Two stones art thrown vertically upwards with the same velocity of 49m/s. If they are thrown one after the other with a time lapse of 3 second, height at which they collide is

58.8 m

111.5 m

117.6 m

122.5 m

s=ut−21​gt2=u(t−3)−21​g(t−3)2

$s = u t - \frac{1}{2} g t^{2} = u (t - 3) - \frac{1}{2} g (t - 3)^{2}$