An aeroplane is travelling horizontally at a speed of 80ms and drops a crate of emergency supplies.To avoid damage,the maximum vertical speed of the crate on landing is 20ms.
a)the aeroplane is travelling at the maximum permitted height.Calc the horizontal distance travelled by the crate after it is released from the aeroplane.
given the speed of 20 m/s
vf^2=2gh
h=20^2/2*9.8= 200/9.8 m
time in air:
hf=hi-4.9t^2
t= sqrt(200/(9.8*4.9))
horizontal distance
distance= 80m/s*timeinairabove
Why did the crate go to a therapist? Because it had a case of "altitude anxiety"!
To calculate the horizontal distance traveled by the crate after it is released from the aeroplane, we can use the formula for horizontal motion:
Distance = Speed * Time
Since the plane is traveling horizontally at a speed of 80 m/s, and the maximum vertical speed of the crate is 20 m/s, we can assume that the horizontal speed of the crate remains constant at 80 m/s.
Let's assume that the crate falls for a time 't' before hitting the ground. During this time, the crate will continue moving horizontally at a speed of 80 m/s. So the horizontal distance traveled by the crate can be calculated as:
Distance = Speed * Time = 80 m/s * t
Now we need to find the time 't', where the maximum vertical speed of the crate is 20 m/s. Since the vertical speed is only due to the force of gravity, we can use the following formula to calculate 't':
Vertical Speed = Acceleration * Time
-20 m/s = -9.8 m/s^2 * t
Solving for 't', we find:
t = 20 m/s / 9.8 m/s^2 ≈ 2.04 s
Now we can substitute this value of 't' back into the equation for distance:
Distance = 80 m/s * 2.04 s ≈ 163.2 m
So, the horizontal distance traveled by the crate after it is released from the aeroplane is approximately 163.2 meters.
To calculate the horizontal distance traveled by the crate after it is released from the airplane, we can use the concept of projectile motion.
Assuming that there is no air resistance, we can break down the motion of the crate into horizontal and vertical components.
Given:
Horizontal speed of the airplane (Vx) = 80 m/s
Vertical speed of the crate (Vy) = -20 m/s (negative since it is moving downwards)
Time of flight (t) = ?
Using the formula for horizontal distance, which is:
Horizontal distance = Vx * t
We need to find the time of flight (t). To do this, we can use the vertical component of motion and time of flight formula:
Vy = g * t
Here, g represents the acceleration due to gravity, which is approximately 9.8 m/s^2.
Substituting the values, we get:
-20 m/s = 9.8 m/s^2 * t
Solving for t gives us:
t = -20 m/s / 9.8 m/s^2 ≈ -2.04 seconds
Please note that the negative sign indicates that it is moving downwards.
Now that we have the time of flight (t), we can calculate the horizontal distance using the horizontal component formula:
Horizontal distance = Vx * t
Substituting the values, we get:
Horizontal distance = 80 m/s * -2.04 s ≈ -162.85 meters
Again, the negative sign indicates that the distance traveled is in the opposite direction of the airplane's motion.
Therefore, when the airplane is traveling at the maximum permitted height, the horizontal distance traveled by the crate after it is released from the airplane is approximately 162.85 meters in the opposite direction of the airplane's motion.
To calculate the horizontal distance traveled by the crate after it is released from the airplane, we need to consider the time of flight of the crate and the horizontal velocity of the airplane.
Since the airplane is traveling horizontally at a speed of 80 m/s, the horizontal velocity of the crate is also 80 m/s.
Given that the maximum vertical speed of the crate on landing is 20 m/s, we can say that the time taken by the crate to reach the ground is the same as the time taken for the airplane to travel the horizontal distance.
To find the time of flight, we can use the equation for vertical motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the crate is dropped and not thrown, the initial velocity, u, is 0 m/s vertically, and the final velocity, v, is 20 m/s downward.
We can rearrange the equation to solve for time:
t = (v - u) / a.
Here, a is the acceleration due to gravity, which is approximately 9.8 m/s^2 downward.
Plugging in the values:
t = (20 - 0) / 9.8
t ≈ 2.04 seconds.
Now, we can find the horizontal distance traveled by the crate using the equation:
s = ut,
where s is the distance traveled, u is the initial horizontal velocity of the crate (which is the same as the horizontal velocity of the airplane), and t is the time of flight.
Plugging in the values:
s = 80 m/s * 2.04 s
s ≈ 163.2 meters.
Therefore, the horizontal distance traveled by the crate after it is released from the airplane is approximately 163.2 meters.