1.A skateboarder moving at 6.15 m/s along a horizontal section of a track that i
ID: 1403156 • Letter: 1
Question
1.A skateboarder moving at 6.15 m/s along a horizontal section of a track that is slanted upward by 46.9 ° above the horizontal at its end, which is 0.544 m above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height H to which she rises above the end of the track.
2.A tennis ball is struck and departs from the racket horizontally with a speed of 29.4 m/s. The ball hits the court at a horizontal distance of 20.7 m from the racket. How far above the court is the tennis ball when it leaves the racket?
3.A 51.8-g golf ball is driven from the tee with an initial speed of 56.7 m/s and rises to a height of 34.4 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point.(b) What is its speed when it is 8.43 m below its highest point?
Someone help me with the confused questions.Thanks a lot
Explanation / Answer
Given,
U = 6.15 m/s
h = o.544m
According to third equation of motion
v^2 - u^2 = -2gh
V^2 = u^2 - 2gh
v = sqrt(u^2 - 2gh)
v = sqrt(6.15^2 - 2*9.8*0.544)
v = 5.21 m/s
Max height of projectile is given by = v^2 sin(theta)^2 / 2g
H = v^2 sin(theta)^2 / 2g
H = (5.21^2 * sin(46.9)^2 )/ 2 * 9.8
H= 0.7383m = 73.83cm
maximum height H to which she rises above the end of the track H= 73.83cm
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