Question

The longest javelin throw at the Olympics covered a horizontal
distance of 104.8 m. The athlete threw the javelin at an angle of
30^{0} above the horizontal from an initial height of 1.50
m above the ground. Ignore air resistance.

a)What was the initial speed of the javelin?

b)What was the maximum height of the javelin?

c)What is the final velocity of the javelin? Give your answer in terms of a magnitude and a direction.

Answer #1

I suppose the angle is 30 Degree

a) Horizontal velocity = vcos30

vertical velocity =vsin30

along vertical

-1.5=vsin30t-4.9t^2

along horizontal

104.8=vcos30 *t

t=104.8/v cos 30

use this t in the first equation

-1.5 = 104.8 tan30-4.9(104.8/vcos30)^2

-1.5-60.51= -71755.86/v^2

v=30.01 m/s

b)Lets get the Max height from the point its being thrown

v^2=u^2+2as along vertical

0= (30.01sin30)^2-2*9.8*s

s=14.76 m

So the max height =14.76+1.5=16.26m

c) Horizontal componenet = 30.01*cos30 =25.989m/s

vertical componenet when it hits the ground

v^2=u^2+2as

v^2=15.08^2+2*9.8*1.5

v=16.025 m/s

Net speed =Sqrt(25.989^2+16.025^2)

v=30.53 m/s

Angle= tan inverse(16.025/25.989)

=31.66 degrees to the horizontal

Kindly check the calculations once

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