Question

an athelete throws a javeline with velocity of 20 meter per sech​

Answer 1

$\huge\tt\fcolorbox{red}{bl}{Answer}$

Explanation:range of javeline,

R=

(u^2)(sin2Φ)/g

i.e.{(20) ^2 }sin 2 * 15 ° / 9.8

=400sin30°/9.8

=(400 * 1/2 )/ 9.8

=20.4 m

Answer 2

Answer:

Explanation:range of javeline,

R=(u^2)(sin20)/g

i.e.{(20) ^2}sin 2 * 15 ° / 9.8

=400sin30°/9.8

=(400 * 1/2 )/ 9.8

= 20.4 m.