A ball's projected at an angle X to the horizontal from a cliff 96 metres high and strikes the ground at a distance of 260 metres away from the cliff.The time of its flight is 7 seconds.If it's assumed that air resistance's negligible,calculate: 1.the initial vertical velocity of the ball, 2.the time to reach the greatest height, 3.the initial horizontal velocity, 4.the angle X,and, 5.the velocity of the projectile on hitting the ground.
S = initial speed
Vi = S sin X
u = S cos X = 260/7 = 37.1 forever (part 3)
v = Vi - g t = S sin X - 9.81 t
h = 96 + Vi t - 4.9 t^2
at 7 seconds, h = 0
0 = 96 + Vi * 7 - 4.9 * 49
7 Vi = 14.41
Vi = 20.6 m/s = S sin X (answer #1)
but S cos X = 260/7 = 37.1
so
tan X = sin X/cos X = 20.6 / 37.1 = 0.555
X = 29 degrees above horizontal (part 4)
S = u/cos x = 37.1 / .875 = 42.4 m/s
max height when v = 0
v = 20.6 - 9.81 t = 0 at top
t = 2.1 seconds to top (part 2)
v at ground after7 seconds
v = 20.6 - 9.81*7 = - 48 m/s
u = 37.1 still
so
velocity at ground = 37.1 i - 48 j (part 5)
onward ....