Question

Two balls are projected from the same point in directions 30° and 60° with respect to the horizontal. What is the ratio of their initial velocities if they (a) attain the same height? (b) have the same range?

Answer 1

Hii dear,

# Given-
θ1=30°
θ2=60°

# Formula-
- Height-
H = u^2.(sinθ)^2 / 2g
- Range-
R = u^2.sin2θ / g

# Calculations-
Let u1 and u2 be the initial velocities for angle 30° and angle 60°.

a) When range is same,
R1 = R2
u1^2.sin60/g = u2^2.sin120/g
u1^2×0.866 = u2^2×0.866
u1 = u2
u1/u2 = 1

b) When heights are same,
H1 = H2
u1^2.(sin30)^2/2g = u2^2.(sin60)^2/2g
u1^2×(0.5)^2 = u2^2×(0.866)^2
0.25×u1^2 = 0.75×u2^2
u1^2 = 3×u2^2
u1/u2 = √3

Hope that helped you...

Answer 2

(a) attain the same height

we know, maximum height , $h_{max}=\frac{u^2sin^2\theta}{2g}$

given, $\theta_1=30^{\circ}$
$\theta_2=60^{\circ}$

$h_1=\frac{u_1^2sin^2\theta_1}{2g}$

= $\frac{u_1^2sin^230^{\circ}}{2g}$

= $\frac{u_1^2}{8g}$......(1)


$h_2=\frac{u_2^2sin^2\theta_2}{2g}$

= $\frac{u_2^2sin^260^{\circ}}{2g}$

= $\frac{3u_2^2}{8g}$

as, $h_1=h_2$

$\frac{u_1^2}{8g}=\frac{3u_2^2}{8g}$

$\frac{u_1}{u_2}=\frac{\sqrt{3}}{1}$


(b) attain the same range.
horizontal range , $R=\frac{u^2sin2\theta}{g}$

$u_1^2sin2\theta_1=u_2^2sin2\theta_2$

$u_1^2sin2(30)=u^2sin2(60)$

$u_1^2\frac{\sqrt{3}}{2}=u_2^2\frac{\sqrt{3}}{2}$

$u_1=u_2$

$\frac{u_1}{u_2}=1$