A cannon ball is fired horizontally with a velocity of 50m/s from the top of a cliff 90m high,after how many seconds will it strike the plane at the foot of the cliff.at what distance from the foot of the cliff will it strike and with what velocity will it strike the ground.

Answer 1

0.5g*t^2 = 90 m.
4.9t^2 = 90
t^2 = 18.37
t = 4.29 s.

Distance = Xo*t = 50m/s * 4.29s=214.3 m.

V = Vo+g*t = 0 + 9.8*4.29 = 42 m/s.

Answer 2

Xo = 50 m/s. = Hor. component.
Y = 42 m/s = Ver. component.
Tot. velocity = sqrt(50^2+42^2)=65.3 m/s

Answer 3

I don't under stand

Answer 4

I don't really understand this too

Answer 5

l don't understand

Answer 6

U=50m/s
H=90m
H=gt^2/2=90=10t^2/2
180=10t^2
t^2=180/10
t^2=√18
t=4.24sec