Question

A shotputter throws the shot with an initial speed of 15.0 m/s at a 37.0 ∘ angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 m above the ground.

Answer 1

Here we have

vertical velocity of shortput = = 9.027 m/s

time taken to reach the ground ,using the kinematic equation we have

putting s= -2.10 m

u = 9.027 m/s

we have

4.9t² - 9.027 t -2.10 = 0

solving the quadratic equation we have

t = 2.051 seconds

now

horizontal velocity =

horizontal distance covered = (horizontal velocity )(time of flight) = 11.98 2.051 = 24.57 m

the horizontal distance traveled by the shot = 24.57 m