Question

In the centennial (on the occasion of its centenary) Olympics held at Atlanta in 1996, the gold medallist hammer thrower threw the hammer to a distance of 19.6m. Assuming this to be the maximum range, calculate the initial speed with which the hammer was thrown. What was the maximum height of the hammer? How long did it remain in the air? Ignore the height of the thrower’s hand above the ground. 

Answer 1

Since we can ignore the height of the thrower’s hand above the ground, the launch point and the point of impact can be taken to be at the same height. We take the origin of the coordinate axes at the launch point. Since the distance covered by the hammer is the range, it is equal to the hammer’s range for θ₀ = 450 . Thus we have from Eqn.(4.7): 

It is given that R = 19.6 m. Putting g = 9.8 ms^–2 we get

The maximum height and time of flight are given by Eqns. (4.5) and (4.6), respectively. Putting the value of v₀ and sin θ₀ in Eqns. (4.5) and (4.6), we get

Now that you have studied some concepts related to projectile motion and their applications, you may like to check your understanding. Solve the following problems.