During a stunt driving show, a physics student drives a car from a parking deck
ID: 1882221 • Letter: D
Question
During a stunt driving show, a physics student drives a car from a parking deck at a vertical height h = 5 m above the ground with an initial horizontal speed v0 = 45 mph, as shown, and lands on the ground at a certain horizontal displacement from the edge of the deck. Assume level ground and free fall conditions.[Take 1 m/s = 2.24 mph and 30.5 cm = 1 foot] (a) What is the horizontal displacement of the car when it lands on the ground? (b) Calculate the speed (in mph) and direction of the velocity of the car as it lands on the ground.
Explanation / Answer
a)
Given Yo=5 m (initial height)
Initial horizontal speed Vox=45 mph
and Voy =0
From
Y=Yo+Voyt-(1/2)gt2
when it lands on ground Y=0
0 =5+0-(1/2)(9.81)t2
t=1.0096 s
Horizontal displacement
X=Voxt =(45*(1/2.24))(1.0096)
X=20.283 m
b)
Vfx =Vox =45 mph
Vfy=Voy-gt =0-9.81*1.0096=-9.9045 m/s
in mph
Vfy =-9.9045*2.24 =-22.186 mph
Speed
V=sqrt[452+(-22.186)2]=50.17 mph
Direction
o=tan-1(-22.186/45)=-26.24o or 153.76o (Counterclock wise from +x-axis)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.