3. A movie stunt man is to run across a rooftop and then horizontally off it, to land on the roof of the next building. The second building's roof is 4.8 m below and 6.2 m distant from the edge of the first building. Before he attempts to make the jump, he wisely does some calculations.
a. If his horizontal velocity is 4.5 m/s, will he safely make the jump?
b. What is the minimum velocity he must have in order to safely make the jump?
c. If his horizontal velocity is only 4.0 m/s, where will the stunt man hit the side of the second
building?
d. If the second building is only 4 m wide, what is the maximum velocity the stunt man can have
without missing the second building altogether?
a. If his horizontal velocity is 4.5 m/s, will he safely make the jump?
b. What is the minimum velocity he must have in order to safely make the jump?
c. If his horizontal velocity is only 4.0 m/s, where will the stunt man hit the side of the second
building?
d. If the second building is only 4 m wide, what is the maximum velocity the stunt man can have
without missing the second building altogether?
-
Time to fall t1 = √[2y/g] = √[2*4.8/9.8] = .99 sec
a) D4.5 = V*t1 = 4.5*.99 = 4.53 m. Answer: No.
b) Vmin = D/t = 6.2/.99 = 6.26 m/s
c) t2 = D/V2 = 6.2/4.0 = 1.55 sec → y = ½g*t2² = 11.77 m. Point of impact = 11.77 - 4.8 = 6.97 m below the top of target building.
d) V3 = D3/t1 = (6.2+4)/.99 = 11.59 m/s max
a) D4.5 = V*t1 = 4.5*.99 = 4.53 m. Answer: No.
b) Vmin = D/t = 6.2/.99 = 6.26 m/s
c) t2 = D/V2 = 6.2/4.0 = 1.55 sec → y = ½g*t2² = 11.77 m. Point of impact = 11.77 - 4.8 = 6.97 m below the top of target building.
d) V3 = D3/t1 = (6.2+4)/.99 = 11.59 m/s max
-
Ignoring air resistance, the horizontal component of his velocity will not diminish. The vertical component of his velocity is Vy = Vyo-gt. Vyo is his initial vertical velocity which I assume to be zero.
Therefore, the moment he takes flight, he is dropping at acceleration g. If he isn't going horizontally fast enough he will drop below the next roof, hit the wall and fall to his death.
A. Vx*t = 6.2m = 4.5t => t = 6.2/4.5 = 1.38s. It will take him 1.38s to reach the next roof. How far will he fall in 1.38s?
d = (1/2)gt^2=0.5*9.8*1.38^2 = 9.3m So by the time he reaches the next building, he is beneath the roof. Oops. Bye!
Calculate the time it takes to fall 4.8m = 4.9t^2 => t=sqrt(4.8/4.9) = 0.99s
Vx* 0.99s = 6.2 => Vx = 6.2/0.99 = 6.26m/s b)
C. For Vx = 4, t=6.2m/4m/s=1.55s
In 1.55s he will fall d=4.9*1.55^2 = 11.77m => 11.77-4.8=6.97m beneath the roof of the next bldg.
D. 6.2+4=10.2m is the horizontal distance he must travel BEFORE falling 4.8m
To fall 4.8m takes t = sqrt[4.8/4.9]=0.99s
How fast must he be going to travel 10.2m in 0.99s? Vx = 10.2/0.99=10.31m/s
10.31m/s*1ft/.3048m*1mile/5280ft*3600s/… = 23.1 miles/hr = 37.1 km/hr
Therefore, the moment he takes flight, he is dropping at acceleration g. If he isn't going horizontally fast enough he will drop below the next roof, hit the wall and fall to his death.
A. Vx*t = 6.2m = 4.5t => t = 6.2/4.5 = 1.38s. It will take him 1.38s to reach the next roof. How far will he fall in 1.38s?
d = (1/2)gt^2=0.5*9.8*1.38^2 = 9.3m So by the time he reaches the next building, he is beneath the roof. Oops. Bye!
Calculate the time it takes to fall 4.8m = 4.9t^2 => t=sqrt(4.8/4.9) = 0.99s
Vx* 0.99s = 6.2 => Vx = 6.2/0.99 = 6.26m/s b)
C. For Vx = 4, t=6.2m/4m/s=1.55s
In 1.55s he will fall d=4.9*1.55^2 = 11.77m => 11.77-4.8=6.97m beneath the roof of the next bldg.
D. 6.2+4=10.2m is the horizontal distance he must travel BEFORE falling 4.8m
To fall 4.8m takes t = sqrt[4.8/4.9]=0.99s
How fast must he be going to travel 10.2m in 0.99s? Vx = 10.2/0.99=10.31m/s
10.31m/s*1ft/.3048m*1mile/5280ft*3600s/… = 23.1 miles/hr = 37.1 km/hr