Question

A stone h = 80m above the ground is released with the first velocity V0 = 0. (g=10 m / s^2 ) 1) How far will the stone travel in the first second? 2) How long will the stone travel in the last 1 second of the movement without hitting the ground?

Answer 1

Answer:

5 m

35 m

Explanation:

From the formula

s = ut + 1 /2 g t²

u = 0  ; t = 1.

s = 1 / 2 x 10 x 1 = 5 m

In first second distance traveled is 5 m .

Time to travel 80 m is t

80 = 0 + .5 x 10 t²

t = 4s

In last one second

s = u + $\frac{2t-1}{2}\times g$

= 0 + (2 x4 -1 )x 10 /2 = 35 m .