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.

See 9:59 AM post.

Plz solve it

To find the time it takes for the cannonball to hit the plane at the foot of the cliff, we need to calculate the time it takes for the cannonball to fall vertically from the top of the cliff to the ground. Since the cannonball is fired horizontally, its horizontal velocity does not change.

Step 1: Calculate the time taken to fall from the top of the cliff to the ground.
Using the formula for vertical displacement under constant acceleration:
s = ut + (1/2)at^2

Here, s = 90m (vertical displacement),
u (initial vertical velocity) = 0 (as the cannonball is fired horizontally),
a (acceleration due to gravity) = -9.8m/s^2 (as acceleration acts downwards),
and t = unknown (time taken).

We can rearrange the formula to solve for time:
90 = 0*t + (1/2)(-9.8)t^2
90 = -4.9t^2

Divide both sides by -4.9:
t^2 = -90/-4.9
t^2 ≈ 18.367

Taking the square root of both sides:
t ≈ √18.367
t ≈ 4.28s

Therefore, it will take approximately 4.28 seconds for the cannonball to fall from the top of the cliff to the ground.

Step 2: Calculate the horizontal distance traveled by the cannonball in 4.28 seconds.
Since the horizontal velocity of the cannonball remains constant at 50m/s throughout its trajectory, we can multiply the velocity by time to find the distance traveled horizontally:
Distance = velocity × time
Distance = 50m/s × 4.28s
Distance ≈ 214m

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

Step 3: Calculate the velocity at which the cannonball strikes the ground.
Since the horizontal velocity of the cannonball remains constant at 50m/s, we only need to find the vertical velocity at which it hits the ground. This can be calculated using the formula:
v = u + at

Here, u (initial vertical velocity) = 0 (as the cannonball is fired horizontally),
a (acceleration due to gravity) = -9.8m/s^2 (as acceleration acts downwards),
and t = 4.28s (time taken to fall).

Using the formula:
v = 0 + (-9.8)(4.28)
v ≈ -42m/s

Therefore, the cannonball will strike the ground with a velocity of approximately -42m/s (in the downward direction).

To solve this problem, we can use the equations of projectile motion. Let's break down the problem step by step to find the answers.

Step 1: Determine the time of flight.
To find the time it takes for the cannonball to strike the plane at the foot of the cliff, we need to find the time it takes for the cannonball to fall to the ground. Since the initial velocity of the cannonball is horizontal, it will not have any vertical acceleration until it starts to fall. Therefore, we can use the equation for vertical displacement:
Δy = V₀y * t + (1/2) * a * t²

Since the cannonball is fired horizontally, the initial vertical velocity (V₀y) is 0. The acceleration (a) due to gravity is -9.8 m/s² (taking downward as negative). The vertical displacement (Δy) is -90 m (negative because it's downward).

Plugging in the known values, the equation becomes:
-90 = 0 * t + (1/2) * (-9.8) * t²

Simplifying the equation:
-90 = (-4.9) * t²

Dividing both sides by -4.9:
t² = 18.37

To solve for t, take the square root of both sides:
t = √18.37
t ≈ 4.28 seconds (rounded to two decimal places)

So, it will take approximately 4.28 seconds for the cannonball to strike the plane at the foot of the cliff.

Step 2: Determine the horizontal distance.
Since the initial horizontal velocity (V₀x) remains constant throughout the motion, we can use the equation:
Δx = V₀x * t

The initial horizontal velocity (V₀x) is 50 m/s, as given. The time (t) is 4.28 seconds, as calculated in step 1.

Plugging in the values:
Δx = 50 * 4.28
Δx ≈ 214 meters (rounded to two decimal places)

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

Step 3: Determine the velocity at impact.
Since the cannonball is only acted upon by gravity in the vertical direction, its vertical velocity (Vy) will increase as it falls. At the moment of impact, the final vertical velocity (Vfy) will equal the initial vertical velocity (V₀y) plus the product of acceleration (a) and time (t).

Vfy = V₀y + a * t

Initial vertical velocity (V₀y) is 0, acceleration (a) is -9.8 m/s², and time (t) is 4.28 seconds.

Plugging in the values:
Vfy = 0 + (-9.8) * 4.28
Vfy ≈ -41.94 m/s (rounded to two decimal places)

The negative sign indicates that the velocity is in the downward direction. Therefore, the cannonball will strike the ground with a velocity of approximately 41.94 m/s in the downward direction.

To summarize:
1. The time it takes for the cannonball to strike the plane at the foot of the cliff is about 4.28 seconds.
2. The cannonball will strike the ground at a horizontal distance of around 214 meters from the foot of the cliff.
3. The velocity at impact is approximately 41.94 m/s downward.