Question

A ball is drop from a height of 420 m. At
the same instant another ball is thrown
up
with an velocity of 720 km/h.
When and where they meet?​

Answer 1

Answer:

t=2.1sec h=398.3m

Explanation:

h=ut-1/2gt2

200t= 1/2×9.8×t2

h= 200t-4.9t2........1

420-h=1/2gt2

420-(200t-4.9t2)=4.9t2

t=2.1 a

h=200t-4.9t2

h=398.3m