Dr. Storr challenged a student to kick a soccerball across a 42, 900.0 mm wide c

Question

Dr. Storr challenged a student to kick a soccerball across a 42, 900.0 mm wide canyon which is at the same height on both sides. The student kicked the ball with all his Herculean strength which was placed at the edge of the canyon and achieved an initial velocity of 20.3 m/s at an angle of 39.0degree. Perform the calculations to see if the ball makes it to the other side. How much short or long of the canyon does it travel? Calculate the time of flight of the ball in seconds. Determine the maximum height of the ball in meters. What is the speed of the ball when it is 5m of the ground?

Explanation / Answer

step;1

Given that

cannon distance =42.9 m

initial velocity of the ball u=20.3 m/s

angle=39 degree

step;2

now we find the time cannon can travell

the time t=[2*42.9/9.8]^1/2=8.8 sec

step;3

now we find the time of flight of the balls

the time of the flight T=2usin(theta)/g=2*20.3*sin(39)/9.8=2.61 sec

now we find the maximum height does the ball reach

the maximum height Hmax=20.3^2sin^2(39)/2*9.8

=8.33 m

step;4

now we find the velocity at distance 5 m

V^2-(20.3*sin(39))^2=2*9.8*5

v^2=163.2+98=261.2

the velocity V=16.2 m/s

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