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3 years ago

A hockey puck sliding on smooth ice at 4 m/s comes to a 1-m-high hill. Will it make it to the top of the hill?

Physics

1 answer:

solniwko [45]3 years ago

3 0

Answer:

No it will not make to the top of the hill

Explanation:

From the question we are told that

The velocity is $v = 4 \ m/s$

The height of the hill is $h = 1 \ m$

Generally from the law of energy conservation we have that

The kinetic energy of the pluck at the level position = The potential energy of the pluck at the maximum height the pluck can get to

$\frac{1}{2} * m * v^2 = m * g * h_{max}$

=> $\frac{1}{2} * v^2 = g * h_{max}$

=> $h_{max} = \frac{0.5 * v^2}{g}$

=> $h_{max} = 0.8163 \ m$

Given that the maximum height which the pluck can get to at this speed given in the question is less than the height of the hill then conclusion will be that the pluck will not make it to the top of the hill

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3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

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a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

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3 years ago