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.

In the vertical direction, the ball is falling under gravity and we know the following:

V0 = 0 m/sec
a = 9.81 m/s
H = 90 m

we can apply the following velocity equation:
Vf^2 = V0^2 + 2 * a * H
Vf^2 = 0 + 2 * 9.81 * 90
=1765.8
Vf = Sqrt(1765.8) = 42.02 m/s
We could apply the velocity equation:
Vf = V0 + at
42 = 0 + 9.81 * t
t = 42/9.81 =4.28 seconds
i.e. the ball hits the ground after 4.28 seconds during which it would have travelled a horizontal distance given by:

distance = V(horizontal) * t
=50 * 4.28 = 214 m
The velocity would be the resultant of both the vertical velocity (42 m/s) and horizontal velocity (50 m/s)

V = sqrt(50^2 + 42^2)= 65.3 m/sec

Why did the cannon ball go on a trip? It wanted to give the plane a little "kick" from the foot of the cliff!

To calculate the time it takes for the cannon ball to strike the plane, we can use the equation for vertical motion:

h = (1/2) * g * t^2

where h is the height (90m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time it takes to strike the plane.

Rearranging the equation, we have:

t = sqrt((2h) / g)

Substituting the given values, we get:

t = sqrt((2 * 90) / 9.8)
t ≈ sqrt(18.37)
t ≈ 4.28 seconds

So it will take approximately 4.28 seconds for the cannon ball to strike the plane.

Now, to find the horizontal distance from the foot of the cliff where it will strike, we can use the equation for horizontal motion:

d = v * t

where d is the distance, v is the horizontal velocity (50 m/s), and t is the time (4.28 seconds).

Substituting the given values, we get:

d = 50 * 4.28
d ≈ 214 meters

So, it will strike the ground at a distance of approximately 214 meters from the foot of the cliff.

As for the velocity at which it will strike the ground, since the only horizontal force acting on the cannon ball is the initial velocity, it will maintain a constant horizontal velocity of 50 m/s until it hits the ground. So it will strike the ground with a velocity of 50 m/s.

To solve this problem, we can use the equations of motion in both the horizontal and vertical directions.

First, let's calculate the time it takes for the cannonball to reach the ground. Since the cannonball is fired horizontally, its initial vertical velocity is zero. We can use the equation:

s = ut + (1/2)at²,

where s is the vertical displacement (90 m), u is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s²), and t is the time taken.

Plugging in the values, we get:

90 = 0 * t + (1/2) * (-9.8) * t²,
90 = (1/2) * (-9.8) * t²,
t² = -90 * 2 / -9.8,
t² = 18.3673,
t ≈ √18.3673,
t ≈ 4.28 seconds.

So, it takes approximately 4.28 seconds for the cannonball to hit the ground.

Next, let's calculate the horizontal distance traveled by the cannonball. Since the initial horizontal velocity is 50 m/s, we can multiply it by the time taken to get the distance:

Distance = Velocity * Time,
Distance = 50 m/s * 4.28 s,
Distance ≈ 214 meters.

Therefore, the cannonball will strike the ground at a distance of approximately 214 meters from the foot of the cliff.

Finally, let's calculate the velocity at which it strikes the ground. Since there is no horizontal acceleration, the horizontal velocity remains constant at 50 m/s. The vertical velocity can be calculated using the equation:

v = u + at,

where v is the final vertical velocity, u is the initial vertical velocity (0 m/s), a is the acceleration due to gravity (-9.8 m/s²), and t is the time taken (4.28 seconds).

Plugging in the values, we get:

v = 0 + (-9.8) * 4.28,
v ≈ -41.944 m/s.

Therefore, the cannonball will strike the ground with a velocity of approximately -41.944 m/s. The negative sign indicates that the velocity is directed downward.