1)A rookie quarterback throws a football with an initial upward velocity compone
ID: 1688818 • Letter: 1
Question
1)A rookie quarterback throws a football with an initial upward velocity component of 15.8 m/s and a horizontal velocity component of 20.3 m/s. Ignore air resistance.a)How much time is required for the football to reach the highest point of the trajectory?
b)How high is this point?
c)How much time (after it is thrown) is required for the football to return to its original level?
d)How far has it traveled horizontally during this time?
2)A man stands on the roof of a building of height 13.2 m and throws a rock with a velocity of magnitude 29.4 m/s at an angle of 31.5^circ above the horizontal. You can ignore air resistance.
a)Calculate the maximum height above the roof reached by the rock.
b)Calculate the magnitude of the velocity of the rock just before it strikes the ground.
c)Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
3)In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below.
a)If the plane is flying in a horizontal path at an altitude of 85.0 m above the ground and with a speed of 66.0 m/s, at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
Explanation / Answer
1. Here a football is projected upwords. The upward velocity component of projection,uy = 15.8 m/s The horizontal velocity component of projection,ux = 20.3 m/s (a) The time taken to football to reach its heighest point of trajectory is given by t = uy/g t = 15.8/9.8 = 1.6 sec (b) The maximum height, hmax = (uy2)/2g = 15.8^2/19.6 = 12.74 m (c) The time taken to reach its original level after thrown is given by t = 2uy/g = 2(15.8)/9.8 = 3.22 sec (d) The horizontal distance travelled by the football is given by R = 2uxuy/g = 2(20.3)(15.8)/9.8 = 65.46 m 2. Height of the building,h = 13.2 m The velocity of projection,u = 29.4 m/s The angle of projection, theeta = 31.5 degrees (a) Maximum height reached by the rock, hmax = u^2sin^?/2g = (29.2)^2sin^2(31.5)/2(9.8) = 11.88 m/s (b) The magnitude of velocity of the rock just before it strikes the ground v^2 = vx^2 + vy^2 vy^2 = u^2 + 2as = (u sin?)^2 + 2g(13.2) = (29.4 sin31.5)^2 + 2(9.8)(13.2) vy = 22.24 m/s vx = u cos? = 29.4cos31.5 = 25.06 m/s v^2 = (25.06)^2 + (22.24)^2 v = 33.5 m/s (c) The horizontal distance travelled by the rock R = u(2h/g)^1/2 R = 29.4(2*13.2/9.8)^1/2 = 48.25 m 3. (a) The height of the plane above the ground, h = 80 m The horizontal velocity of the plane,u = 66 m/s ? The horizontal distance from the target should the pilot release the canister R = u(2h/g)^1/2 = 66(2*80/9.8)^1/2 = 266.7 mRelated Questions
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