Asked by Temi

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.

Answers

Answered by Henry
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.


Answered by Henry
Xo = 50 m/s. = Hor. component.
Y = 42 m/s = Ver. component.
Tot. velocity = sqrt(50^2+42^2)=65.3 m/s
Answered by vickky
I don't under stand
Answered by Shaf
I don't really understand this too
Answered by Anonymous
l don't understand
Answered by Ifunanya ugorji
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
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions