In an exhibition, you win a prize if you toss a coin into a small dish placed. The dish is on a sheep $2.1 m$ away at a height $h$ from the hand. The coin is tossed into the dish if it is projected at velocity of $6.4 m s^{- 1}$ at an angle $60^{\circ}$ . Find $h$ .

1.2 m

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1.35 m

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1.5 m

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1.65 m

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Solution

The correct option is D $1.5 m$
horizontal component of the velocity $v_{x} = v c o s θ = v c o s 60^{o} = \frac{v}{2}$
vertical component $v_{y} = v s i n 60^{o} = \frac{\sqrt{3} v}{2}$
time taken to cover 2.1m is
$S = v_{x} t$
$2.1 = 3.2 t$

$t = \frac{21}{32} t$
therefore vertical height h
$h = v_{y} t - \frac{1}{2} g t^{2}$

$h = \frac{\sqrt{3} v}{2} \times (\frac{21}{32}) - \frac{1}{2} g (\frac{21}{32})^{2} = 1.5 m$

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In an exhibition, you win a prize if you toss a coin into a small dish placed. The dish is on a sheep 2.1 m away at a height h from the hand. The coin is tossed into the dish if it is projected at velocity of 6.4 ms−1 at an angle 60∘. Find h.

The correct option is D 1.5 mhorizontal component of the velocity vx=vcosθ=vcos60o=v2vertical component vy=vsin60o=√3v2time taken to cover 2.1m is S=vxt2.1=3.2tt=2132ttherefore vertical height hh=vyt−12gt2h=√3v2×(2132)−12g(2132)2=1.5m

In an exhibition, you win a prize if you toss a coin into a small dish placed. The dish is on a sheep $2.1 m$ away at a height $h$ from the hand. The coin is tossed into the dish if it is projected at velocity of $6.4 m s^{- 1}$ at an angle $60^{\circ}$ . Find $h$ .