A projectile is fired with an initial speed of 120m.s' at an angle of 60' above the horizontal from the top of a 50m high cliff. (a) determine the maximum height(above the cliff) reached by the projectile (b) how long it takes to fall to the ground from the max. height?. Please give me some help i'm confuse.
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)
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
posted by Nadir
thanks a lot Nadirposted by Mykmoloko