A ski jumper starts from rest from point A at the top of a hill that is a height h1 above point B at the bottom of the hill. The

skier and skis have a combined mass of 80 kg. The skier slides down the hill and then up a ramp and is launched into the air at point C that is a height of 10 m above the ground. The skier reaches point C traveling at 42ms.Find his speed at the bottom of the hill which is 10m below the top.

Physics

You might be interested in

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0

serg [3582]

Since you've completed parts a and b, I will tackle part c.
For part C
To respond to this question, we must identify the zeros of the velocity function:
$v(t)=0.04t^3-0.06t^2$
This polynomial can be factored:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Finding the zeros now becomes straightforward since the function equals zero when any factor is zero.
$t^2=0;\\ 0.04t-0.06=0$
By solving these equations, we identify our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle remains stationary at t=0 and t=3/2.
For part D
We must discover when the velocity function exceeds zero. We will utilize its factored form.
We will assess when each factor is greater than zero and compile the findings in the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
From the table, it's evident that our function is positive when $- \infty < t$ and t>3/2.
This indicates the interval during which the particle moves forward.
For part E
The distance traveled can be represented as:
$s(t)=0.01t^4 - 0.02t^3$
We simply substitute t=12 to calculate the total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
For part F
Acceleration is defined as the rate at which velocity changes.
We determine acceleration by deriving the velocity function concerning time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find the acceleration at 1 second, we substitute t=1s into the previous equation:
$a(1)=0.12-0.12=0$

7 0

2 months ago

The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV

Softa [3030]

For thorough details and necessary calculations, please refer to the attachment.

6 0

2 months ago

A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

ValentinkaMS [3465]

Since the motorcycle was at a speed of 36 m/s prior to braking, that marks the initial velocity.
$u = 36 {ms}^{ - 1}$
The motorcycle skidded 23m before coming to a full stop, indicating that
$s = 23m$
As it has ceased motion, the final velocity is zero.

$v = 0{ms}^{ - 1}$
We can apply the 'suvat' formula relevant to linear motion.

${v}^{2} = {u}^{2} + 2as$
Substituting the aforementioned values allows us to find,

${0}^{2} = {36}^{2} + 2a(23)$

$0 = 1296+ 46a$
$46a = - 1296$
$a = - 28.2 {ms}^{ - 2}$

We can apply the formula
$v = u + at$
to calculate the time required for the motorcycle to stop.

$0 = 36 + - 28.2t$
$- 36 = - 28.2t$
$t = 1.3s$

8 0

3 months ago