Question

a hiker stands on the edge of a cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 M per second neglecting resistance find the time taken by the stone to reach the ground and their speed with which it hits the ground take gravity is equal to 9.8 metre per second square

Answer 1

1.) time taken to reach the ground.

We will use the following formula.

S = ut + 0.5gt²

S = 15t + 0.5 × 9.8t²

S = 15t + 4.9t²

490 = 15t + 4.9t²

4.9t² + 15t - 490 = 0

Using quadratic equation formula :

t = ( -15 +/-√225 + 9604) / 9.8

t = (-15 +/- 99.14) / 9.8

t = 8.59 seconds

2.) Range = ut

Range = 15 × 8.59 = 128.85 m

= 128.85 M

3.) V² = U² + 2gs

V² = 225 + 9604

V² = 9829

V = 99.14 m/s