Question

An athlete throws a javelin to a maximum distance of 80 m. How long is it in air

and to what height does it rise. Take g = 9.8 m/s2​

Answer 1

Answer: Height= 20m.

Explanation:

Let the speed by which the stone is thrown be  u making an angle θ with the horizontal.

∴ Horizontal range     R=u² sin2θ/g

For maximum horizontal range ,θ=45  

⟹    Rmax =    u²sin90 /g

⇒        80   = u²/g

⇒        80   =u²/9.8

⇒         u²   = 784

Let maximum height it attains be  H

∴   H= u²sin²θ/ 2g

where   θ=45  

⇒ H = 784×1/2  / 2×9.8

⇒ H = 20m.

Hope this help you !