Need help with a physics problem (involving formulas - yr 11 stuff) (1 Viewer)

[Supaweak]

Hello, the question is:

A ball is thrown upwards with velocity of 9.8m/s from a cliff of height 995.1m. How high will it rise above the cliff and what will be its velocity when it reaches the ground beneath the cliff?


The answer is exactly 140m/s and you use the formula V^2 = U^2 +2as

I've tried it a few times and I keep getting 140.34m/s or something similar, my teacher did the question on the board a while ago but I didn't get a chance to fully take it in

any help will be appreciated, thanks!

 

[Drongoski]

Hello, the question is:

A ball is thrown upwards with velocity of 9.8m/s from a cliff of height 995.1m. How high will it rise above the cliff and what will be its velocity when it reaches the ground beneath the cliff?


The answer is exactly 140m/s and you use the formula V^2 = U^2 +2as

I've tried it a few times and I keep getting 140.34m/s or something similar, my teacher did the question on the board a while ago but I didn't get a chance to fully take it in

any help will be appreciated, thanks![/quote



You have used the correct formula: v^2 = u^2 + 2as

i) on the way up, taking 'up' as positive direction: v = 0, u = 9.8, a = - 9.8

so you get: 0^2 = 9.8^2 + 2 x (-9.8) x s
so : s = 4.9 m

ii) on the way down, taking 'down' as positive direction: u = 0, a = 9.8, v = ? and
s = (995 + 4.9) m = 1000 m (4.9 m = height above cliff just worked out above)

so: v^2 = 0^2 + 2 x 9.8 x 1000 = 140^2
Therefore v = 140 m/sec

Do you get it now Supaweak ??

 

[Supaweak]

ahh thanks for the reply i get it now thanks!