Physics, 20.01.2020 06:31 Angelanova69134

In a scene in an action movie, a stuntman jumps from the top of one
building to the top of another building 4.0 m away. after a running start,
he leaps at a velocity of 5.0 m/s at an angle of 15° with respect to the flat
roof. will he make it to the other roof, which is 2.5 m lower than the
building he jumps from?

Answers: 3

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Answer from: howey74

Yes, the stuntman can jump from one building to the other building.

Further explanation:

The stuntman jump from one building inclined at some angle to the other traversing a parabolic path.

Given:

The velocity of stuntman is 5m/s.

The distance between the buildings is 4m.

The difference in the height of the buildings is 2.5m.

The angle of inclination is ${15^ \circ }$ .

Concept used:

When stuntman jumps from the top of one building to the top of other building he start running first then he jump at a velocity of 5m/s inclined at an angle of ${15^ \circ }$ from the horizontal of first building.

The velocity of stuntman has two components ${v_x}$ and ${v_y}$ , respectively in the X-direction and in the Y-direction.

The expression for the distance in horizontal direction is given as.

$x = \left( {v\cos \theta } \right)t$

Rearrange the above expression for time.

$t = \dfrac{x}{{\left( {v\cos \theta } \right)}}$ …… (1)

Here, t is the time of flight, v is the velocity of object and $\theta$ is the angle of inclination.

The expression for the distance in Y-direction is given by the second equation of motion.

$y = ut + \frac{1}{2}\left( { - g} \right){t^2}$

Here, u is the velocity in Y-direction and (–g) is the acceleration due to gravity directed in the downward direction.

The expression for the component of velocity in Y-direction is given as.

$u = v\sin \theta$

Substitute $v\sin\theta$ for u in the above expression.

$y = \left( {v\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}$ …… (2)

Substitute 5m/s for v, 4m for x and ${15^ \circ }$ for in equation (1).

$\begin{aligned}t&=\frac{{4\,{\text{m}}}}{{\left({\left( {5\,{\text{m/s}}}\right)\left( {\cos {{15}^ \circ }}\right)}\right)}}\\&=0.828\,{\text{s}}\\\end{aligned}$

Substitute $0.828\,{\text{s}}$ for t, 5m/s for v, $9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}$ for g and ${15^ \circ }$ for $\theta$ in equation (2).

$\begin{aligned}y&=\left( {\left( {5\,{\text{m/s}}} \right)\sin \left( {{{15}^ \circ }} \right)} \right)\left( {0.828\,{\text{s}}} \right) + \frac{1}{2}\left( { - 9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right){\left( {0.828\,{\text{s}}} \right)^2}\\&=1.071 - 3.36 \\&=- 2.29\,{\text{m}}\\\end{aligned}$

Thus, the stuntman can jump from one building to another because its vertical distance is less than the difference in height of buildings.

Learn more:

1. Motion under force link.

2. Projection of ball link.

3. Conservation of momentum link.

Answer Details:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords:

Force, motion, parabolic path, acceleration due to gravity, time of flight, angle of inclination, inclined plane, buildings, height difference, cliff, horizontal direction, vertical direction, 2.29 m, 2.3m, 0.828sec.

Answer from: Quest

be subject to a force andchange its motion

Answer from: Quest

Isso9

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