Jack tries to toss his baseball up onto the roof of a garage. At the instant whe
ID: 1463317 • Letter: J
Question
Jack tries to toss his baseball up onto the roof of a garage. At the instant when the ball leaves Jack's hand, the ball is traveling with a velocity of 9.00 m/s directed 70.0 degrees above horizontal. At this instant, the ball is located 1.00 m above the ground, 2.00 m below the level of the garage roof, at a horizontal distance of 4.00 m from the near wall of the garage.
Part A) The ball would land on the garage roof at a horizontal distance of _____ m from its launch point.
Part B) Just before the ball strikes the garage roof, the ball's speed would be equal to _____ m/s.
Please show work!
Explanation / Answer
A)
vo = 9 m/s
voy = vo*sin(70) = 9*sin(70) = 8.46 m/s
vox = vo*cos(70) = 9*cos(70) = 3.08 m/s
y =2 m
x = 4 m
let t is the time to land on the roof,
Apply, y = voy*t - 0.5*g*t^2
2 = 8.46*t - 4.9*t^2
4.9*t^2 - 8.46*t + 2 = 0
on sloving the above equation
t = 1.4438 s
so, landing distance from launch point = vox*t
= 3.08*1.4438
= 4.447 m <<<<<<<<----------Answer
b) at landing pioint,
vy = voy - g*t
= 8.46 - 9.8*1.4438
= -5.69 m/s
vx = vox = 3.08 m/s
so, v = sqrt(vx^2 + vy^2)
= sqrt(3.08^2 + 5.69^2)
= 6.47 m/s <<<<<<<<----------Answer
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