Question

It's been a great day of new, frictionless snow. Julie starts at the top of the 60∘ slope shown in the figure. At the bottom, a circular arc carries her through a 90∘ turn, and she then launches off a 3.0-m-high ramp. What horizontal distance will she travel before touching down?

Answer 1

She travels approximately 40m before touching down.

Equating inital and final kinetic and potential energies,

$\frac{1}{2}$m$v_{i} ^{2}$ + mg$h_{i}$ = $\frac{1}{2}$m$v_{f} ^{2}$ + mg$h_{f}$

Solving for velocity, at the end of the ramp, v = sqrt($\sqrt{2g(y1 - y2)}$)

Now, velocity as she is going off the ramp should be

$v_{f}$= sqrt($\sqrt{2g(25-3}$ = 20.8 m/s

Now,  y =$y_{0}$ + $v_{0}$t - $\frac{1}{2}$g$t^{2}$    
            = 3 + 20.8 sin (60) - 4.9$t^{2}$
            = 3 + 18.01t - 4.9$t^{2}$ 
            t = 3.83
Now, distance x = speed*time 
                            =  vt = $v_{0}$cos(θ)t
                            = 20.8 * cos(60) * 3.83
                            = 40 m

Answer 2

Julie will travel a distance of almost $\fbox{\bf{42.7\text{\bf{ m}}}}$ before she touches the ground.

Further Explanation:

Since Julie starts to travel down the incline from a height of $25\,{\text{m}}$. Therefore, Julie has some potential energy initially and as she travels down the inclined, she attains some speed and the potential energy of Julie gets converted into the kinetic energy.

Apply the conservation of energy at the initial point and the final point of the ramp.

$\fbox{\begin\\mg{h_i}+\frac{1}{2}mv_i^2=mg{h_f}+\frac{1}{2}mv^2\end{minispace}}$

Here, ${h_i}$ is the initial height of the inclined, ${h_f}$ is the final height from where Julie takes off, ${v_i}$ is the initial velocity of Julie and ${v}$ is the final velocity of Julie.

Initially, Julie starts from rest. So, the initial kinetic energy will be zero.

Rearrange the above expression for the final velocity of Julie at the take off point $3\,{\text{m}}$ above the ground.

$v=\sqrt{2g\left({{h_i}-{h_f}}\right)}$

Substitute $25\,{\text{m}}$ for ${h_i}$ and $3\,{\text{m}}$ for ${h_f}$ and $9.81\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}$ for $g$ in above expression.

$\begin{aligned}v&=\sqrt{2\left({9.81}\right)\left({25-3}\right)}\,{{\text{m}} \mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\&=20.8\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

As soon as Julie takes off from this point above the ground, she moves under the action of the gravity and follows a projectile path.

Now we calculate the time taken by Julie to reach the ground in the vertical direction using the second equation of motion.

$S={u_y}t-\frac{1}{2}g{t^2}$

Here, $S$ is the displacement of Julie in vertical direction, $u$ is the initial velocity of Julie as she takes off from the ramp and $t$ is the time taken by her to reach the ground.

Since the direction of displacement of Julie is in downward direction so, the displacement will be negative and the speed of Julie in the vertical direction will be $u\sin 30^\circ$.

Substitute $-3\,{\text{m}}$ for $S$, $20.8\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for $u$ and $9.81\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$ for $g$ in above expression.

$\begin{aligned}\\-3&=(20.8\sin30^\circ)t-\dfrac{1}{2}\left( {9.81}\right){t^2}\\0&=4.905{t^2}-10.4t-3\\t&=2.37\,{\text{s}}\\\end{aligned}}$

The speed of Julie in the horizontal direction will be $u\cos 30^\circ$.

So, the horizontal distance covered by Julie after taking off from the ramp will be.

$\begin{aligned}{\text{distance}}& = {\text{speed}} \times {\text{time}} \\&= 20.8\cos 30^\circ  \times 2.37\,{\text{m}} \\&= 4{\text{2}}{\text{.69}}\,{\text{m}}\\&\approx {\text{42.7}}\,{\text{m}} \\ \end{aligned}$

Thus, Julie will cover a distance of almost $\fbox{\bf{42.7\text{\bf{ m}}}}$ after taking off from the ramp.

Learn More:

1.  A ball falling under the acceleration due to gravity https://brainly.com/question/10934170

2.  The stress developed in a wire https://brainly.com/question/12985068

3. Translational kinetic energy of a gas molecule https://brainly.com/question/9078768

Answer Details:

Grade: High School

Subject: Physics

Chapter: Conservation of energy

Keywords:

Julie, ramp, 60 degree, 60, 25 m, takes off, 3 m, circular arc, carries through, kinetic energy, potential, horizontal distance, displacement, slope.