an aeroplane is travelling horizontally at a speed of 80ms and drops a crate of emergency supplies.To avoid damage,the max vertical speed of the crate on landing is 20ms.

a)calc d max height of the aeroplane when the crate is dropped
b)calc d time taken for the crate to reach the ground from this height.

v = at

20 = 9.8 t
t = 2.04s
so, it can't fall for more than 2.04 seconds, or it will be going too fast

so, how far will it fall in 2.04 s?
s = 1/2 at^2
s = 4.9 * 4.16
s = 20.38m

Man, that's some low flyer!

Very helpful

Using V2 = U2 + 2gh

U=0
V2 = 2gh
h = V2/2g
h = 400/19.62
h = 20.39m

Thank you very much. this was very helpful.

Shouldn't we like need to find the total velocity first since horizontal speed is 80m/s and vertical speed is 20m/s?

a) Well, if the crate is dropped from the airplane, it will experience two types of motion: horizontal and vertical. The horizontal speed of the airplane doesn't affect the maximum height, so we can ignore it for now. We need to find the maximum height, which means the vertical speed at that point will be zero.

Since the crate's vertical speed can't exceed 20 m/s, its initial vertical speed will be 20 m/s (assuming it's thrown downwards), and it will eventually reach zero velocity at the maximum height.

Using the equation v^2 = u^2 + 2as, where v is the final velocity (zero in this case), u is the initial velocity (20 m/s downwards), a is acceleration (due to gravity, approximately 9.8 m/s^2), and s is the distance traveled vertically, we can calculate the maximum height (s).

Rearranging the equation, s = (v^2 - u^2) / (2a), we get:
s = (0 - 20^2) / (2 * -9.8)

Now, don't you go falling through this complicated equation like a crate from an airplane! After crunching the numbers, we find that the maximum height of the aeroplane when the crate is dropped is approximately 20.4 meters.

b) To calculate the time it takes for the crate to reach the ground from this maximum height, we can use another equation from kinematics. The equation is v = u + at, where v is the final velocity, u is the initial velocity, a is acceleration, and t is time.

In this case, the final velocity is the speed at which the crate hits the ground (which is 0 m/s because it comes to a stop there). The initial velocity is the vertical speed with which the crate was thrown downwards (20 m/s), and the acceleration is due to gravity (-9.8 m/s^2).

Plugging these values into the equation, we get:
0 = 20 + (-9.8)t

Now, it's time to do some calculations! Solving this equation, we find that t is approximately 2.04 seconds.

So, the time taken for the crate to reach the ground from the maximum height is around 2.04 seconds. I hope that answer didn't drop too many jaws!

To solve this problem, we'll need to use the equations of motion and understand the motion of the crate when it's dropped from the airplane.

a) Calculating the maximum height of the airplane when the crate is dropped:
To determine the maximum height of the airplane, we need to find the time it takes for the crate to reach the ground. Given that the maximum vertical speed of the crate on landing is 20 m/s, we can use the following equation:

v² = u² + 2as

Where:
- v is the final velocity (20 m/s),
- u is the initial velocity (0 m/s since the crate starts from rest),
- a is the acceleration (due to gravity, approximately -9.8 m/s² since the crate is moving upward),
- s is the displacement (which is the maximum height we want to find).

Rearranging the equation, we have:

s = (v² - u²) / (2a)

Substituting the given values:

s = (20² - 0²) / (2 * -9.8)

Calculating this expression, we find:

s ≈ 20.41 meters

Hence, the maximum height of the airplane when the crate is dropped is approximately 20.41 meters.

b) Calculating the time taken for the crate to reach the ground from this height:
To find the time taken for the crate to reach the ground, we can use the equation of motion:

v = u + at

Where:
- v is the final velocity (0 m/s at the ground),
- u is the initial velocity (20 m/s),
- a is the acceleration (due to gravity, approximately -9.8 m/s²),
- t is the time taken.

Rearranging the equation, we have:

t = (v - u) / a

Substituting the given values:

t = (0 - 20) / (-9.8)

Calculating this expression, we find:

t ≈ 2.04 seconds

Therefore, the time taken for the crate to reach the ground from the maximum height is approximately 2.04 seconds.