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.
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
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
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!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.