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tiny-mole [99]
3 years ago
7

A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

Physics
1 answer:
aleksley [76]3 years ago
3 0
Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
                            d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
                          d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters. 
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A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at
vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

g=-9.8 m/s^2 (acceleration of gravity)

The initial vertical velocity is

u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s

where the negative sign means the direction is downward.

The vertical position of the ball is given by

y(t) = h + u_y t + \frac{1}{2}gt^2

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

0 = 4.50 -1.96t-4.9t^2

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s

where u = 3.05 m/s is the initial speed and \theta the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

d=v_x t

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

d=(2.34)(0.78)=1.83 m

7 0
3 years ago
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