Question

A shotputter throws the shot (mass = 7.3 kg) with an initial speed of 14.4 m/s at a 34.0º 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

Answer:

22.2 m

Explanation:

m = 7.3 kg

initial speed, u = 14.4 m/s

angle of projection, θ = 34°

vertical height, h = 2.1 m

Use second equation of motion for vertical direction , let t be the time taken

$h = u_{y}t + \frac{1}{2}gt^{2}$

- 2.1 = 14.4 Sin 34 t - 0.5 x 9.8 x t²

- 2.1 = 8 t - 4.9 t²

4.9 t² - 8 t - 2.1 = 0

$t=\frac{8\pm \sqrt{8^{2}+ 4\times 2.1\times 4.9}}{9.8}$

t = 1.86 second

So, the horizontal distance, d = horizontal velocity x time

d = 14.4 x Cos 34 x 1.86

d = 22.2 m

Thus, the horizontal distance traveled is 22.2 m .