# Physics Homework Help?

A ball is thrown vertically upwards in the air with an initial speed of 10 m/s and from a point that is 4.8 m above the ground. How long does it take for the ball to hit the ground? Take g = 10 m/s2 and ignore air resistance.

### 2 Answers

Relevance

- oubaasLv 78 months ago
in two steps :

Δh = Vo^2/2g = 10^2/20 = 5 m

tup = Vo/g = 10/10 = 1.000 sec

tdown = √ 2(h+Δh)/g = √ (5+4.8)*2/10 = 1.400 sec

t = tup + tdown = 1.00+1.40 = 2.400 sec

in one step

(0-4.8) = Vo*t-g/2*t^2

-4.8-10t+5t^2 = 0

t = (10+√ 10^2+96)/10 = (10+14)/10 = 24/10 = 2.400 sec

- FiremanLv 78 months ago
Given h = 4.8 m, u = -10 m/s & g = 10 m/s^2, let the ball hit the ground after t sec,

By s = ut + 1/2gt^2

=>4.8 = - 10t + 1/2 x 10 x t^2

=>5t^2 - 10t - 4.8 = 0

=>t = [-(-10) +/- √{(-10)^2 - 4 x 5 x (-4.8)}]/[2 x 5]

=>t = [10 +/- 14]/10

=>t = 2.4 sec

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