Question

A man on a motorcycle plans to make a jump as shown in the figure. If he leaves the ramp with a speed of 40.0 m/s and has a speed of 37.7 m/s at the top of his trajectory, determine his maximum height (n) (in m) above the end of the ramp. Ignore friction and air resistance. m

Answer 1

of 40 Solution Moliseyele leaves the samp with a speed m/sec. Motorcycle has a speed of 37.7 msec at the top of trajectory. So initial velocity u= 40 m/sec 37.7 m/sec u=4omface usino Let the angle 0 , which is the angle between initial velocity and horizonatal. Resobure initial velocity in vertical and horizontal component Vertical component = 40 Sino mpec Horizontal component = u Coso = 40 cose am/sec There is no accelaration in hosizontal direction, so the horiztal component of initial velocity semains constant through out the motion Therefore e = cos" ( 972 ) = 19:52 40 Coso = 37.7

motion is Formula of Maximum Maximum Height, h height of projectile u2 Sin ²0 h 2g h = (40)² Sin (1952) 2x9.8 u=40 m/sec 0 = 19.52 Ig= acceleration due = 1.8m h = 1600 x 0.11 rec to gravity -3 289.8 h = 8.9 m h = 8.9 m ~ 9 m Hence Maximum height above the ramp is 8.9 m of an approx.