HELP WITH TWO PHYSICS QUESTIONS PLEASE :)

[Oliveme]

1) A student, standing on the platform at a railway station, notices that the first 2 carriages of an arriving train pass her in 2s, and the next 2 in 2.4s. The train is decelerating uniformly. Each carriage is 20m long. When the train stops, the student is opposite the last carriage. How many carriages are there in the train?

2) A ball is to be kicked so that, at the highest point of its path, it just clears a horizontal cross-bar on a pair of goal-posts. The ground is level and the cross-bar is 2.5m high. The ball is kicked from ground level with an initial velocity of 8 m/s.

a) calculate the angle of projection of the ball and the distance of the point where the ball was kicked from the goal-line.
b) Also calculate the horizontal velocity of the ball as it passes over the cross-bar.
c) for how long is the ball in the air before it reaches the ground on the far side of the cross bar?

Thanks very much.

 

[The Godfather]

1) We can find the initial velocity first.
The first two carriages passed by in 2 s
Each carriage is 20 m long
Average initial velocity = 40 / 2 = 20 m/s

The second two passed by in 2.4 s
Average velocity = 40 / 2.4 = 16.67 m/s

From these two velocities we can find the average decelleration...
Average decelleration = 20 - 16.67 / 4.4 = 0.76 m / s^2

Using
v^2 = u^2 - 2as
where v = final velocity; u = initial velcoity; a = acceleration; s = distance
then
2as = u^2
s = u^2 / 2a = 400 / 2 * 0.76 = 264.3 m

If each carriage = 20m
then number of carriages = 264.3 / 20 = 13.2
but we can't have 0.2 of a carriage, so this 0.2 must be part of the last carriage making the total number of carriages 14.

2)
a) From the equations of projectile motion we have

vertical height = u sinѲ t - (1/2) g t^2
vertical velocity = u sinѲ - gt

At its highest point, the vertical velocity is momentarily zero. We can use this fact to work out how long it takes to get to its max height..

0 = u sinѲ - gt

t = u sinѲ / g

We can now plug this value of time into the equation for its height

vertical height = u sinѲ( u sinѲ/g) - (1/2) g (u sinѲ/g)^2

vertical height = (u sinѲ)^2 / 2g

But we are told that at its max height the ball just clears the post which is 2.5 m high. So

2.5 = u^2 sin^2 (Ѳ) / 2g

sin^2(Ѳ) = 2g * 2.5 / u^2 = 2 * 9.8 * 2.5 / 64 = 0.766

sin Ѳ = √0.766 = 0.875
Ѳ = arc sin (0.875) = 61.04 degrees.

b) The horizontal velocity remains constant throughout the flight and is given by

u cosѲ t
where t = total travel time
The total travel is just twice the time required to reach its max height. So
total travel time = 2 u sinѲ / g

Horizontal velocity = u cosѲ 2 u sinѲ / g = u^2 2 cosѲ sinѲ / g

2 cosѲ sinѲ = sin(2Ѳ)

Horizontal velocity = u^2 sin(2Ѳ) / g = 64 * 0.841 / 9.8 = 5.5 m / s

c) total travel time = 2 u sinѲ / g = 2 * 8 * 0.875 / 9.8 = 1.43 s