Question

Obtain the expression for maximum height and horizontao range of projectile

Answer 1

➡️$\underline{Horizontal\: range:-}$The total distance covered by the body in projectile is called horizontal range.

Horizontal range = Horizontal component of velocity × time of flight

$or, \: R = ucos \theta \: \times \frac{2usin \theta }{g } \\ \\ or, \: R = \frac{u {}^{2} sin2 \theta }{g}$

➡️$\underline{Maximum\: horizontal\: range:-}$ The horizontal range will be maximum when sin2θ is maximum.

$sin2 \theta = 1 = sin90° \\ or, \: 2 \theta = 90° \\ \theta = 45°$

➡️$\underline{Maximum\: Height:-}$The vertical component of velocity at maximum height is 0. So,

$v_{y} {}^{2} = (usin \alpha ) {}^{2} - 2gh \\ 0 = u {}^{2} sin \theta - 2gh \\ h = \frac{u {}^{2}sin {}^{2} \theta }{2g}$

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