0 votes
in Physics by (73.7k points)

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. 

1 Answer

+1 vote
by (75.6k points)
selected by
Best answer

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.

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.