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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.

Verified Answer :

MissPhiladelphia

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

Answer:

22.7 m

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:

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

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