the maximum height attained by the projectile is increased by 10% by increasing speed of projection , without changing the angle of projection. calculate the percentage in horizontal range
Answers
Answered by
2
Initial max. height is, h = (u2 sin2θ)/2g
Time of flight, t = 2u sinθ/g
=> t2 = 4u2 sin2θ/g2
=> t2 = (8/g)[ (u2 sin2θ)/2g] = (8/g)h
Therefore,
T2/t2 = H/h
Now, new height, H = h + (10/100)h = (11/10)h
=> H/h = 11/10
And,
T2/t2 = 11/10
=> T/t = 1.05
=> T/t – 1 = 0.05
=> (T-t)t = 0.05
=> (T-t)t × 100 = 5%
keep asking:+
Follow us
Time of flight, t = 2u sinθ/g
=> t2 = 4u2 sin2θ/g2
=> t2 = (8/g)[ (u2 sin2θ)/2g] = (8/g)h
Therefore,
T2/t2 = H/h
Now, new height, H = h + (10/100)h = (11/10)h
=> H/h = 11/10
And,
T2/t2 = 11/10
=> T/t = 1.05
=> T/t – 1 = 0.05
=> (T-t)t = 0.05
=> (T-t)t × 100 = 5%
keep asking:+
Follow us
Similar questions