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.
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.
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
I don't under stand
To find the time it takes for the cannonball to strike the plane at the foot of the cliff, we can use the equation of motion for vertical motion:
h = ut + (1/2) * g * t^2
where:
h = height of the cliff (90m)
u = initial vertical velocity (0 m/s since it is fired horizontally)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Since the initial vertical velocity is 0 m/s, the equation simplifies to:
h = (1/2) * g * t^2
Plugging in the values:
90 = (1/2) * 9.8 * t^2
Now, let's solve for t:
180 = 9.8 * t^2
t^2 = 180 / 9.8
t^2 = 18.37
t = √18.37
t ≈ 4.28 seconds (rounded to two decimal places)
Therefore, it takes approximately 4.28 seconds for the cannonball to strike the plane at the foot of the cliff.
To find the horizontal distance from the foot of the cliff where the cannonball will strike, we can use the equation for horizontal motion:
d = v * t
where:
d = horizontal distance
v = horizontal velocity (50 m/s)
t = time (4.28 seconds)
Plugging in the values:
d = 50 * 4.28
d ≈ 214 meters
Therefore, the cannonball will strike the ground at a horizontal distance of approximately 214 meters from the foot of the cliff.
Since the cannonball was fired horizontally, its vertical velocity remains 0 m/s. However, its horizontal velocity remains constant at 50 m/s until it strikes the ground.
I don't really understand this too
l don't understand
Xo = 50 m/s. = Hor. component.
Y = 42 m/s = Ver. component.
Tot. velocity = sqrt(50^2+42^2)=65.3 m/s