First slide

Projection Under uniform Acceleration

Question

An object is projected with a velocity of 20 ms^-1 making an angle of 45⁰ with horizontal. The equation for the trajectory is h = Ax - Bx² where 'h' is height, x is horizontal distance, A and B are constants. The ratio A: B is (g = 10 ms^-2)

Moderate

A

1:5
B

5:1
C

1:40
D

40 :1

Solution

For a projectile, if 'h' be its height at a distance x from the point of projection, then h = x tanα(1 - x/R) where α is the angle of projection and R is range. Comparing with h = Ax - Bx²,
A = tanα =1 and
$\fn_phv \large B = \frac{1}{R} = \frac{g}{{{u^2}{{\sin }2\alpha}}$

$\therefore \;\frac{A}{B} = \frac{tan\alpha }{{g/u^{2}sin2\alpha }}$

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