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ID: 2051362 • Letter: H
Question
http://ef.engr.utk.edu/c/sys/13/assess/paramimagev2.php?n=/www/c/ef151-2012-01/img/assess/trajectory-hill.png&p_HT=85,31,257,0&p_V0=49,155,50,0&p_SL=2,319,300,0&p_ANG=60,169,96,0Peyton throws a ball from a tower towards the side of hill as shown. The height above the ground at which he releases the ball, the speed and angle at which he releases the ball, and the slope of the hill are given in the diagram.
A) What is the maximum height that the ball reaches? (ft above x-axis)
B) How long does it take the ball to reach its maximum height? (sec)
C) What is the x coordinate where the ball hits the hill? (ft)
D) What is the y coordinate where the ball hits the hill? (ft)
E) What is the speed of the ball when it hits the hill? (ft/s)
Explanation / Answer
A)&B) at the maximum height Vy = 0
49sin60-g*t = 0
t = (49*3/2)/(32.15) = 1.32 s
it takes 1.32 s to reach maximum height
vertical distance travelled
x = 49*sin60*t-1/2*g*t^2 = 49*3/2*1.32-1/2*32.15*(1.32)^2 = 28 ft
max height reached = 85+x = 85+28 = 113 ft
C)&D)
Horizontal distance travelled before hitting hill = 49*cos60*t = 49/2*t = x
Vertical distance travelled before hitting hill = 49*sin60*t-1/2*g*t^2 = 49*3/2*t-1/2*32.15*t^2
y = 85+(49*3/2*t-1/2*32.15*t^2)
we have been given ratio of
horizontal/vertical at the hill x/y = 2/1
[49/2*t]/(85+[49*3/2*t-1/2*32.15*t^2]) = 2
solving for t gives us
Horizontal distance travelled before hitting hill = 49*cos60*t = 49/2*t = 83.85566 ft
Vertical distance travelled before hitting hill = 49*sin60*t-1/2*g*t^2 = 49*3/2*t-1/2*32.15*t^2 = -43.07 ft
E)
horizontal speed = 49*cos60 = 49/2 = 24.5 ft/sec
verical speed = 49*sin60-g*t = 49*3/2-32.15*3.42268 = - 67.604 ft/sec
speed = (24.5^2+67.604^2) = 71.9 ft/sec
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