A Slun biker jumps across water by Jumping between two symmetric ramps that are

Question

A Slun biker jumps across water by Jumping between two symmetric ramps that are separated by a distance D as shown below. Assume the biker takes off exactly at the end of the first ramp and lands at the top of the second ramp (travels a distance D in the horizontal direction D in the horizontal dorection). Each ramp makes an angle theta - 25.0 degree with the horizontal. The biker leaves the ramp with a speed of vo = 6.7 m/s a height h = 5.0 m above the water. Assume that air resistance can be neglected. How long is the biker in the air? What is the horizontal distance D? What is the maximum height, H, reached by biker above the top of the water?

Explanation / Answer

Here ,

a)

time of flight = 2*v*sin(theta)/g

time of flight = 2 * 6.7 * sin(25)/9.8

time of flight = 0.58 s

the time of flight is 0.58 s

b)

horizontal distance , D = t * v* cos(25)

horizontal distance , D = 0.58 * 6.7 * cos(25)

horizontal distance , D = 3.51 m

c)

Now, height H is

H = 5 + (v*sin(25))^2/(2g)

H = 5 + (6.7 * sin(25))^2/19.6

H = 5.41 m

the maximum height H is 5.41 m

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