Question

A long jumper goes into a jump at an angle of 20.0° above the horizontal and at a speed of 9.9 m/s. What is his time of flight (the time the jumper is in the air), in units of seconds?

Answer 1

Answer:

T = 0.69 s

Explanation:

given,

speed of the long jumper = 9.9 m/s

angle of the jump = 20°

time of flight calculation = ?                            

using equation of time of flight

$T = \dfrac{2\ u \ sin \theta }{g}$                

$T = \dfrac{2\times 9.9 \times sin 20^0}{9.8}$              

T = 0.69 s                                                      

so, the time of flight of the jumper is equal to T = 0.69 s