Question

A shot-putter throws the shot with an initial speed of 15.5m/s at a 34 degree angle to the horizontal. calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.

Answer 1

22.7 m

Step-by-step explanation:

Given,

Initial speed of shot, u =15.5 m/s

Angle at which the shot was thrown is 34°

Horizontal distance covered by projectile is given by the formula:

`R = \frac {u^2 sin 2\theta}{g}\\ R = (15.5^2 sin 68°)/(9.8) =22.7 m`

Answer 2

this can be solve by using the formula
x = (v^2) (sin 2a) / gwhere v is the intial velocitya is the angle of trajectoryg is the acceleration due to gravity (9.8 m/s2)
x = (15.5^2) (sin 2(34) / 9.8x = 22.73 m