A projectile is fired so to secure max. Horizontal range. If its flight time is 10sec find velocity of projection
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T = 2usin∅/g
10 = 2usin∅/g
5 = usin∅/g ...(i)
When ∅ = 45°, max horizontal range is secured, therefore
5 = usin(45°)/g
5 = u/g√2
u = 5g√2
u = 5(9.8)√2
u = 49√2
u = 49(1.41)
u = 69.09
The velocity is 69.09m/s
10 = 2usin∅/g
5 = usin∅/g ...(i)
When ∅ = 45°, max horizontal range is secured, therefore
5 = usin(45°)/g
5 = u/g√2
u = 5g√2
u = 5(9.8)√2
u = 49√2
u = 49(1.41)
u = 69.09
The velocity is 69.09m/s
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Explanation:
a projectile is fired so as to secure the maximum horizontal range its time of flight is 10s find the velocity of its projection g=9.8m/s square
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