Question

HELP PLZ! A parachutist jumps out of an aeroplane and accelerates uniformly at 9.8 ms-2 for ten seconds. a. How far does she travel? b. What is her final velocity?

Answer 1

Answer:

a.) 490m

b.) 98m/s

Explanation:

Given that the

Acceleration g = 9.8 m/s^2

Time = 10s

Since the parachutist jumps out of an aeroplane. The parachutist jumped out from rest. Initial velocity U is therefore equal to zero. That is,

U = 0

Distance covered = height H

The height can be calculated by using second equation of motion

H = Ut + 1/2gt^2

Substitute g and t into the formula

H = 1/2 × 9.8 × 10^2

H = 490 m

Therefore, she travels as far as 490 m

b.) Her final velocity can be calculated by using third equation of motion

V^2 = U^2 + 2gH

Substitute g and H into the formula.

Remember that U = 0

V^2 = 2 × 9.8 × 490

V^2 = 9604

V = sqrt (9604)

V = 98 m/s

Therefore, her final velocity is 98 m/s