A javelin is thrown with a speed of 25 m per second. If the maximum height reached by it is 6 m, what is its angle of projection, duration of flight and range?
Answers
Answer:
Let the angle of projection to the horizontal be x degrees.
Then the vertical component of the velocity will be 25 sinx
Using v^2 = u^2 + 2as, with
v = 0 (at the maximum height)
u = 25 sinx
a = -9.8
s = 6
we get: 25^2 (sinx)^2 - 19.6 x 6 = 0
This solves to give (sinx)^2 = 19.6 x 6 /625
Evaluating, taking the square root and inverse sin gives x = 25.7 degrees.
To find the duration of flight, use s = ut + 0.5 at^2 vertically, with:
u = 25sin25.7
a = -9.8
s = 0 (the vertical displacement when the projectile ends its flight)
Substituting: 25t sin25.7 - 4.9 t^2 = 0
Factorise: t(25 sin25.7 - 4.9 t) = 0
t= 0 (the starting position) or t = (25 sin25.7 )/4.9 = 2.21 seconds
This assumes that the javelin is thrown from ground level. If not, you would have to use s = -h, where h is the height that it is thrown from (not given). The resulting quadratic equation would have to be solved using the quadratic formula.
To find the range, use s = ut + 0.5 at^2 in the horizontal direction, with:
u = 25 cos 25.7 (the horizontal component of the velocity)
t = 2.21 (from above)
a = 0 (no force acting on the javelin in the horizontal direction)
Substituting gives:
s = (25 cos 25.7) x 2.21 = 49.78 metres
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Explanation:
Let the angle of projection to the horizontal be x degrees.
Then the vertical component of the velocity will be 25 sinx
Using v^2 = u^2 + 2as, with
v = 0 (at the maximum height)
u = 25 sinx
a = -9.8
s = 6
we get: 25^2 (sinx)^2 - 19.6 x 6 = 0
This solves to give (sinx)^2 = 19.6 x 6 /625
Evaluating, taking the square root and inverse sin gives x = 25.7 degrees.
To find the duration of flight, use s = ut + 0.5 at^2 vertically, with:
u = 25sin25.7
a = -9.8
s = 0 (the vertical displacement when the projectile ends its flight)
Substituting: 25t sin25.7 - 4.9 t^2 = 0
Factorise: t(25 sin25.7 - 4.9 t) = 0
t= 0 (the starting position) or t = (25 sin25.7 )/4.9 = 2.21 seconds
This assumes that the javelin is thrown from ground level. If not, you would have to use s = -h, where h is the height that it is thrown from (not given). The resulting quadratic equation would have to be solved using the quadratic formula.
To find the range, use s = ut + 0.5 at^2 in the horizontal direction, with:
u = 25 cos 25.7 (the horizontal component of the velocity)
t = 2.21 (from above)
a = 0 (no force acting on the javelin in the horizontal direction)
Substituting gives:
s = (25 cos 25.7) x 2.21 = 49.78 metres
425 views · View 1 Upvoter