A shot putter puts a shot through a horizontal distance 12.56 m. At the instant

Question

A shot putter puts a shot through a horizontal distance 12.56 m. At the instant when the athlete releases the short, the elevation of the shot is 1.71 m above the ground level. The angle of release measured from the horizontal is 300.

Determine the speed, in m s-1, at which the athlete releases the shot. Please show working. Give your answer to 3 decimal places and take g = 9.81 ms-2.

A shot putter puts a shot through a horizontal distance 12.56 m. At the instant when the athlete releases the short, the elevation of the shot is 1.71 m above the ground level. The angle of release measured from the horizontal is 30 degree. Determine the speed, in m s^-1, at which the athlete releases the shot. Please show working. Give your answer to 3 decimal places and take g = 9.81 ms^-2.

Explanation / Answer

Here ,

initially , speed be v

theta = 30 degree

then , when the shot hit ground , assuming initial position as origin

x = 12.56 m

y = -1.71 m

trajetory of projectile ,

y = x*tan(theta) - g*x^2/2(v*cos(theta))^2

-1.71 = 12.56 * tan(30) - 9.8 * 12.56^2 /(2*(v*cos(30))^2)

solving for v

v = 10.72 m/s

the initial speed is 10.72 m/s

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