Four balls A, B, C and D are projected with the same speed making angles of 20 deg, 30 deg, 45 deg and 80 deg with the horizontal. Which ball will strike the ground at nearest point?

1 Answer
Nov 26, 2017

R(\theta_0=80^o) is the minimum.

Explanation:

Range of the Projectile: \qquad R= v_0^2/g\sin2\theta_0

When the speed of projection v_0 is the same, R \propto \sin2\theta_0

\theta_0 = 20^o; \qquad \sin2\theta_0 = 0.6428;
\theta_0 = 30^o; \qquad \sin2\theta_0 = 0.8660;
\theta_0 = 45^o; \qquad \sin2\theta_0 = 1.0000; Maximum
\theta_0 = 80^o; \qquad \sin2\theta_0 = 0.3420; Minimum

R_{max} = R(\theta_0=45);
Among the given angles the minimum range happens to be for a projection angle of \theta_0=80^o.