Consider a rock that is thrown off a bridge of height 37 m at an angle ? = 24° w

Question

Consider a rock that is thrown off a bridge of height 37 m at an angle ? = 24° with respect to the horizontal as shown in the figure below. 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 m/s
direction °

Explanation / Answer

(a) Let the rock attain its maximum height in t1 sec
By v = u - gt
0 = [Uy] - gt1
t1 = usin/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 + 37 = 38.22 m
Let the stone take t2 sec to fall H meter
By s = ut + 1/2gt^2
38.22 = 0 + 1/2 x 9.8 x t2^2
t2 = 7.8
t2 2.79s
Thus the total time to land (T) = t1 + t2 = 0.50 + 2.79 = 3.29 sec
(d) By R = [Ux] x T
R = ucos x T
R = 12 x cos24* x 3.29
R = 36.07 m
(e) Let the angle of landing is and the magnitude of velocity is v m/s
By v = u + gt
Vy = 0 + 9.8 x t2
Vy = 9.8 x 2.79 = 27.34 m/s
& Vx = Ux = 10.96 m/s
v = [(Vx)^2 + (Vy)^2]
v = [(10.96)^2 + (27.34)^2]
v = 29.45 m/s
By tan = Vy/Vx = 27.34/10.96 = 2.49 tan68*
= 68*

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