Question

In a javelin throw an athlete launches the javelin at a speed of 29.0 m/s at an angle of 36ᴼ above the horizontal. On the upward part of the trajectory the velocity vector still points above the horizontal however at a decreasing angle. How much time is required for the angle to be reduced to 18ᴼ

Answer 1

given u = 29 m/s and = 36⁰

during the projectile motion horizontal component of the velocity is remains same and is equals to ucos

hence u_x = u cos36⁰ and u_y = u sin36⁰

suppose after t seconds angle is reduced to 18⁰

at this time v_x = u_x = u cos36⁰

and v_y = u_y - gt = u sin36⁰ - gt

tan18⁰ = v_y / v_x

0.3249 =  (u sin36⁰ - gt) /  u cos36⁰

= u tan36⁰  - ( gt / u cos36⁰)

gt / u cos36⁰ = 29 x 0.7265 - 0.3249

= 20.7436

t = 20.7436 x u cos36⁰ / g

= 20.7436 x 29 x 0.8090 /9.81

= 49.6 s