in the olympic javelin event, a javelin is released at a 39 degree angle above t

Question

in the olympic javelin event, a javelin is released at a 39 degree angle above the horizontal at a height of 1.77 meters above the ground.
a) the javelin travels 88.1m horizontally before returning to its original height. how fast must the javelin be thrown to cover this distance?
in the olympic javelin event, a javelin is released at a 39 degree angle above the horizontal at a height of 1.77 meters above the ground.
a) the javelin travels 88.1m horizontally before returning to its original height. how fast must the javelin be thrown to cover this distance?

a) the javelin travels 88.1m horizontally before returning to its original height. how fast must the javelin be thrown to cover this distance?

Explanation / Answer

Given,

theta = 39 deg ; R = 1.77 m ; D = 88.1 m

we know that

R = v0x T

R = v0 cos(theta) T

T = R/v0 cos(theta)

also, T = 2 x sqrt (2H/g)

H = v0y^2/2 g

T = 2 x sqrt (2 x v0y^2/2g)

R/v0 cos(theta) = 2 v0y sqrt(2/2g)

R/v0 cos(theta) = 2 v0 sin(theta) (0.32)

2 v0^2 sin(theta) cos(theta) = R/0.32 = 88.1/0.32 = 275.31

2 v0 sin(2theta) = 275.31

v0^2 = 275.31/2sin(2 x 39) => v0 = 11.86 m/s

Hence, v0 = 11.86 m/s

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