Answer :
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 [tex]{15^ \circ }[/tex].
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 [tex]{15^ \circ }[/tex] from the horizontal of first building.
The velocity of stuntman has two components [tex]{v_x}[/tex] and [tex]{v_y}[/tex], respectively in the X-direction and in the Y-direction.
The expression for the distance in horizontal direction is given as.
[tex]x = \left( {v\cos \theta } \right)t[/tex]
Rearrange the above expression for time.
[tex]t = \dfrac{x}{{\left( {v\cos \theta } \right)}}[/tex] …… (1)
Here, t is the time of flight, v is the velocity of object and [tex]\theta[/tex] is the angle of inclination.
The expression for the distance in Y-direction is given by the second equation of motion.
[tex]y = ut + \frac{1}{2}\left( { - g} \right){t^2}[/tex]
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.
[tex]u = v\sin \theta[/tex]
Substitute [tex]v\sin\theta[/tex] for u in the above expression.
[tex]y = \left( {v\sin \theta } \right)t + \frac{1}{2}\left( { - g} \right){t^2}[/tex] …… (2)
Substitute 5m/s for v, 4m for x and [tex]{15^ \circ }[/tex] for in equation (1).
[tex]\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}[/tex]
Substitute [tex]0.828\,{\text{s}}[/tex] for t, 5m/s for v, [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for g and [tex]{15^ \circ }[/tex] for [tex]\theta[/tex] in equation (2).
[tex]\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}[/tex]
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.
To know whether he will make it to the other roof that is 2.5m lower, we can say that;
He will make it to the other roof safely because the height he jumped which is 2.3 m is lesser than the roof height of 2.5 m from where he jumped.
We are given;
Velocity; v = 5 m/s
Distance jumped from the top of one building to the top of another building; x = 4 m
Angle made with respect to flat roof; θ = 15°
The formula for the horizontal distance is;
x = t(v cos θ)
Let us find the time taken;
t = x/(v cos θ)
Plugging in the relevant values gives;
t = 4/(5 cos 15)
t = 0.83 seconds
Now, to find the height we will use newtons equation of motion;
y = vt sin θ - ¹/₂gt²
Plugging in the relevant values gives;
y = (5 × 0.83 × sin 15) - ¹/₂(9.8 × 0.83²)
y = 1.074 - 3.376
y ≈ -2.3 m
This height is less than 2.5 m and as such we will conclude that he will make it to the other roof safely.
Read more at; https://brainly.com/question/14006339