First slide

Projectile motion

Question

A body is thrown with the velocity v₀ at an angle of $θ$ to the horizon. Determine v₀ in ms^-1 if the maximum height attained by the body is 5 m and at the highest point of its trajectory the radius of curvature is r = 3 m. Neglect air resistance. [Use $\sqrt{80}$ as 9] $Take g = 10 {ms}^{- 2}$

Moderate

Solution

Radius of curvature at the highest point
$r = \frac{v^{2}}{g} \Rightarrow r = \frac{{(v_{0} \cos θ)}^{2}}{g}$
$v_{0} \cos θ = \sqrt{gr}$ …(i)
Maximum height, $h = \frac{v_{0}^{2} \sin^{2} θ}{g} \Rightarrow v_{0} \sin θ = \sqrt{gh}$ …(ii)
By using Equations (i) and (ii), we get
$v_{0}^{2} \sin^{2} θ + v_{0}^{2} \cos^{2} θ = gh + gr = g (h + r)$
$v_{0} = \sqrt{g (h + r)}$
$\Rightarrow v_{0} = \sqrt{10 (5 + 3)} = 9 {ms}^{- 1}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App