A ball is launched from a 300 m cliff and lands 380 m away from the base of the cliff in 9 seconds. Calculate the initial speed and the angle of the ball when it was launched.
Answers
Explanation:
The initial velocity can be broken into its horizontal and vertical components. Assuming that there is no air resistance, the magnitude of the horizontal component remains constant.
Let u(hz) and u(vr) be the horizontal and vertical components of the initial velocity, respectively, ie.
u(hz) = ucos(a) where a is the angle to the horizontal at which the ball is launched from, and u(vr) = usin(a).
We are given that the ball travels 380m from the cliff in 9 seconds, so we can derive an expression for the horizontal component:
u(hz) = 380/9 = ucos(a)
u = 380/9cos(a) (1)
Also, the ball's vertical displacement is -300m and its acceleration is the acceleration due to gravity ie. -9.8ms^-2, and this takes 9 seconds.
Using the kinematics equation s = ut +1/2(at^2), we get:
-300 = 9u(vr) - 4.9 x 81
u(vr) = (4.9 x 81 - 300)/9 = usin(a)
u = 96.9/9sin(a) (2)
Both (1) and (2) equal u, so equate the two:
96.9/9sin(a) = 380/9cos(a)
96.9cos(a) = 380sin(a)
tan(a) = 96.9/380
a = arctan(96.9/380)
Substitute this into (1):
u = 380/9cos(arctan(96.9/380))
Calculating this:
a = 14 degrees and 31 minutes
u = 43.6 m/s
Therefore, the ball is launched at 43.6 m/s, 14 degrees and 31 minutes above the horizontal.