Question

a projectile is fired horizontally with a velocity of 98m/s from the top of a hill 490 m high.find the time taken to reach the ground , The distance of the target from the foot of hill and The velocity with which the particle hits the ground

Answer 1

1.) Time taken to reach the ground:

S = ut + 0.5gt²

U = 98

g = 10

490 = 98t + 5t²

5t² + 98t - 490 = 0

t = ( -98 +/- √9604 + 9800) / 10

t = (-98 -/+ 139.30) / 10

t = 4.13 seconds.

2.) Range = ut


Range = 98 × 4.13 = 404.74 m

3.) Final velocity:

V² = U² + 2gs

V² = 9604 + 9800

V² = 19404

V = 139.30 m/s

Answer 2

Answer:

Here is the easiest way to solve this question

Explanation:

See the attachment below.