Consider a rock that is thrown off a bridge of height 27 m at an angle θ = 24° with respect to the horizontal as shown in the figure above. If the initial speed the rock is thrown is 12 m/s, find the following quantities:
(a) The time it takes the rock to reach its maximum height.
s
(b) The maximum height reached by the rock.
m
(c) The time at which the rock lands.
s
(d) The place where the rock lands.
m
(e) The velocity of the rock (magnitude and direction) just before it lands.
Magnitude Direction
m/s °
a) Let the rock attain its maximum height in t1 sec
=>By v = u - gt
=>0 = [Uy] - gt1
=>t1 = usinθ/g
=>t1 = [12 x sin24*]/9.8
=>t1 ≈ 0.50 sec
(b) Let the rock attain h meter in t1 sec
=>By v^2 = u^2 - 2gh
=>0 = [Uy]^2 - 2gh
=>h = [12 x sin24*]^2/[2 x 9.8]
=>h = 1.22 m
(c) Total height (H) of the stone fro ground = h + 27 = 28.22 m
Let the stone take t2 sec to fall H meter
=>By s = ut + 1/2gt^2
=>28.22 = 0 + 1/2 x 9.8 x t2^2
=>t2 = √5.76
=>t2 ≈ 2.4sec
Thus the total time to land (T) = t1 + t2 = 0.50 + 2.4 = 2.90 sec
(d) By R = [Ux] x T
=>R = ucosθ x T
=>R = 12 x cos24* x 2.90
=>R = 31.79 m
(e) Let the angle of landing is α and the magnitude of velocity is v m/s
=>By v = u + gt
=>Vy = 0 + 9.8 x t2
=>Vy = 9.8 x 2.4 = 23.52 m/s
& Vx = Ux = 10.96 m/s
=>v = √[(Vx)^2 + (Vy)^2]
=>v = √[(10.96)^2 + (23.52)^2]
=>v = 25.95 m/s
By tanα = Vy/Vx = 23.52/10.96 = 2.15 ≈ tan65*
=>α = 65*
B C E are wrong what am I doing wrong?
1 answer