Posted by Mykmoloko on Wednesday, February 9, 2011 at 10:01am.
1st you find the vertical component of its speed which = (120)sin(60)=103.92 m/s
Gravity's pull will decrease its vertical speed by G (=9.8 m/s^2)
Therefore the vertical velocity of the projectile = 103.92 - (9.8)*t (whee t is time in seconds)
It will reach maximum height when the velocity is 0 (because then it will start falling)
So 103.92 - (9.8)*t = 0 when t = 10.6 seconds (maximum height is reached when t = 10.6 seconds)
to find the height we integrate the vertical velocity eq' (103.92 - (9.8)*t) and we get ( (103.92)*t - (4.9)*t^2 + C ) (note: Since at t = 0, the height = (103.92)*(0) - (4.9)*(0)^2 + C = 0, therefor C = 0. If we assume at t = 0 our height is 50 (because of the cliff, then C = 50 but here we will use the cliff as our reference point therefore C = 0)
we sub in t = 10.6 and we get the maximum height above the cliff = (103.92)*(10.6) - (4.9)*(10.6)^2 = 550.99 meters above the cliff (or 600.99 meters above ground)
Part b:
We can use the vertical velocity equation we used before (103.92)*t - (4.9)*t^2 + C
But here we will put C = 50 because the ground will be our reference point.
Therefore (103.92)*t - (4.9)*t^2 + 50 = 0
Solving this eq' we get t = -0.47 seconds or t = 21.68 seconds, since we cant have negative time we take t = 21.68 seconds
At max height t was 10.6 Seconds, Then time it took from max height to ground = 21.68 - 10.6 = 11.08 seconds
thanks a lot Nadir