First slide

Projection Under uniform Acceleration

Question

A person can throw a stone to a maximum height of h metre. The maximum distance to which he can throw the stone is:

Moderate

A

h
B

$\frac{h}{2}$
C

2h
D

3h

Solution

Height H is given by
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
When $θ = 90^{0}, H = H_{\max} = \frac{u^{2}}{2 g} = h$
Range is given by: R = $\frac{u^{2} \sin 2 θ}{g}$
Now $R_{\max} = \frac{u^{2}}{g} (when θ = 45^{0}) = 2 h$

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