Question
Two particles are projected upwards from level ground with the same initial velocity $\mathrm{v}_{0}$ at two different angles of projection with horizontal, such that their horizontal ranges are the same. The ratio of the heights of their highest points will be(let $\theta_{1}=$ angle of projection of the first particle with horizontal)(a) $\frac{1}{\cos \theta_{1}}$(b) $\sin \theta_{1}$(c) $\tan ^{2} \theta_{1}$(d) $\cos \theta_{1}$
Step 1
Step 1: The maximum height reached by a projectile is given by the formula $H = \frac{v_{0}^{2}\sin^{2}\theta}{2g}$, where $v_{0}$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity. Show more…
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Two particles are projected in air with speed $v_{0}$ at angles $\theta_{1}$ and $\theta_{2}$ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (a) angle of projection : $\theta_{1}>\theta_{2}$ (b) time of flight : $T_{1}>T_{2}$ (c) horizontal range : $R_{1}>R_{2}$ (d) total energy : $U_{1}>U_{2}$
Projectile Motion
Round 2
A particle is projected from the ground with an initial speed of $\mathrm{v}$ at an angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is a. $\frac{v}{2} \sqrt{1+2 \cos ^{2} \theta}$ b. $\frac{v}{2} \sqrt{1+2 \cos ^{2} \theta}$ c. $\frac{v}{2} \sqrt{1+3 \cos ^{2} \theta}$ d. $v \cos \theta$
A particle is projected from the ground with an initial speed of $v$ at an angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is (a) $\frac{v}{2} \sqrt{1+2 \cos ^{2} \theta}$ (b) $\frac{v}{2} \sqrt{1+\cos ^{2} \theta}$ (c) $\frac{v}{2} \sqrt{1+3 \cos ^{2} \theta}$ (d) $v \cos \theta$
Round 1
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