In a scene in an action movie, a stunt man
jumps from the top of one building to the
top of another building 4.9 m away. After a
running start, he leaps at an angle of 17
◦
with
respect to the flat roof while traveling at a
speed of 5.8 m/s.
The acceleration of gravity is 9
.
81 m
/
s
2
.
To determine if he will make it to the other
roof, which is 1.4 m shorter than the build-
ing from which he jumps, find his vertical
displacement upon reaching the front edge of
the lower building with respect to the taller
building.
Answer in units of m
Vo = 5.8m/s[17o]
Xo = 5.8*cos17 = 5.55 m/s.
Yo = 5.8*sin17 = 1.70 m/s.
Y = Yo + g*t = 0 @ max. ht.
Tr = (Y-Yo)/g = (0-1.70)/-9.8 = 0.173 s.
= Rise time.
h=(Y^2-Yo^2)/2g = (0-1.7^2)/-19.6=0.147 m.
h = 1.4 + 0.147 = 1.55 m Above shorter
bldg.
h = 0.5g*t^2 = 1.55 m.
4.9*t^2 = 1.55
t^2 = 0.316
Tf = 0.562 s. = Fall time or time to fall to roof of shorter bldg.
Dx = Xo * (Tr+Tf)
Dx = 5.55m/s * (0.173+0.562) = 4.08 m.
= His Hor. distance from bldg. #1.
Therefore, he will fall short of bldg. #2:
4.9 - 4.08 = 0.82 m. Short.
To solve this problem, we will need to break down the motion of the stuntman into horizontal and vertical components.
Let's start by calculating the horizontal displacement. The stuntman jumps a distance of 4.9 m horizontally. Since there is no horizontal acceleration (assuming air resistance is negligible), the horizontal component of the velocity remains constant throughout the jump. Therefore, the horizontal displacement is simply the product of the horizontal component of the velocity and the time of flight.
To find the time of flight, we can use the vertical component of the velocity. The initial vertical velocity is given as 5.8 m/s, and the angle of 17 degrees also provides information about the vertical motion. Let's separate the initial velocity into vertical and horizontal components:
Vertical component: V_y = V_initial * sin(angle)
Horizontal component: V_x = V_initial * cos(angle)
Using these equations, we can find the vertical component of the velocity: V_y = 5.8 m/s * sin(17 degrees).
Next, we can find the time of flight. The time it takes for the stuntman to reach the lower building is the same as the time it takes for him to travel horizontally. We can use the horizontal displacement and the horizontal component of the velocity to find the time of flight:
Time of flight = Horizontal displacement / Horizontal velocity
= 4.9 m / (5.8 m/s * cos(17 degrees))
Once we have the time of flight, we can find the vertical displacement upon reaching the front edge of the lower building. The vertical displacement is given by the equation:
Vertical displacement = Initial vertical velocity * time + (1/2) * acceleration due to gravity * time^2
However, since the stuntman starts and lands at the same height, the vertical displacement is equal to the height difference between the two buildings, which is given as 1.4 m.
Finally, we can calculate the vertical displacement by substituting the values into the equation:
1.4 m = (5.8 m/s * sin(17 degrees) * time) + (1/2) * (9.81 m/s^2) * time^2.
Here, we have an equation with one unknown, time. We can solve this equation using algebraic methods or use numerical methods to find the value of time. Once we have the value of time, we can substitute it back into the equation for vertical displacement to obtain the final answer.