Question

(ii) Two stones are thrown towards each other simultaneously one vertically downwards from the top of the tower of height 360m with velocity 10m/s and other from the ground vertically upwards with velocity 50m/s .Find the time and the distance from the ground at which they collide. Take g=10m/s^(2) ​

Answer 1

Explanation:

height covered by the projectile from the ground

h=ut−

2

1

gt

2

h=100t−

2

1

×9.8×t

2

h=100t–4.9t

2

…(1)

Distance covered by the projectile thrown down in time t will be, since it falls down from the restu=0

300–h=

2

1

gt

2

300–(100t–4.9t

2

)=4.9t

2

…byequation(1)

300=100t,t=3s

Thus particles meet each other at 3s

Height at which particles meet

h=100t–4.9t

2

h=100×3–4.9×9

h=255.9m