A slingshot is used to launch a stone horizontally from the top of a 20.0 meter

Question

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away. At what speed was the stone launched? What is the speed and angle of impact? A canon ball fired horizontally from a cliff has a velocity directed at 60 degree below horizontal when it hits the ground 3.0 seconds later. How high is the cliff? How far from the base of the cliff does the canon ball land? A ball rolls with a speed of 2.0 m/s across a level table that is 1.0 m above the floor. Upon reaching the edge of the table, it follows a parabolic path to the floor. How far along the floor is the landing spot from the table? A baseball player leads off the game and hits a long home run. The ball leaves the bat at an angle of 30.0 degree from the horizontal with a velocity of 40.0 m/s. How far will it travel in the air? A golfer is teeing off on a 170.0 m long par 3 hole. The ball leaves with a velocity of 40.0 m/s at 50.0 degree to the horizontal. Assuming that she hits the ball on a direct path to the hole, how far from the hole will the ball land (no bounces or rolls)? What is the maximum height of the ball in its flight?

Explanation / Answer

1)

along horizantal

x = v*t=====> t = x/v

along vertical

y = 0.5*g*t^2

y = 0.5*g*x^2/v^2

20 = 0.5*9.8*36^2/v^2

v= 17.82 m/s

(b)

time t = sqrt(2y/g) = sqrt((2*20)/9.81) = 2.1 s

along vertical

vy = voy + gt

vy = 0 + 9.81*2.1 = 20.601 m/s

along horzizantal vx = vox = 17.82 m/s

speed = sqrt(20.601^2 + 17.82^2) = 27.24 m/s

angle = tan^-1(vy/vx) = 40.8 below horizantal

+++++++++++++++

(2)

along vertical

y = 0.5*g*t^2

y = 0.5*9.81*3^2

y = 44.145 m

-----

(b)

along vertical

vy = voy + ay*t

vy = gt = 9.81*3 = 29.43 m/s

tan60 = vy/vx

sqrt3 = 29.43/vx

vx = 17 m/s

x = vx*t = 17*3 =51 m

+++++++++++++++

3)

x = v*sqrt(2y/g)

x = 2*sqrt((2*1)/9.81)

x = 0.9 m

++++++++++

4)

range = v^2*(sin(2theta))/g

R = 40^2*sin60/9.81 = 141.25 m

++++++++++

5)

R = v^2*(sin(2theta))/g

R = 160.6 m

R < 170 m

the ball will land at a distance of 10 m

(b)

H = v^2*(sintheta)^2/2g

H = (40^2*(sin50)^2)/(2*9.81) = 47.8 m

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