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 https://brainly.com/question/6125929.
2. Projection of ball https://brainly.com/question/11023695.
3. Conservation of momentum https://brainly.com/question/9484203.
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.
