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In a javelin throw an athlete launches the javelin at a speed of 29.0 m/s at...

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

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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 ux = u cos360 and uy = u sin360

suppose after t seconds angle is reduced to 180

at this time vx = ux = u cos360

and vy = uy - gt = u sin360 - gt

tan180 = vy / vx

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

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