Question

A ball is thrown with a velocity of 7 root 2 m/s at an angle of 45 degree with the horizontal. It just clears two vertical poles of height 90 cm. Find the separation between the poles. Take g =9.8

Answer 1

Answer on Question #60387-Physics-Mechanics-Relativity

A ball is thrown with a velocity of 7 root $2\mathrm{m / s}$ at an angle of 45 degree with the horizontal. It just clears two vertical poles of height $90~\mathrm{cm}$ . Find the separation between the poles. Take $\mathbf{g} = 9.8$

Solution

Let $t$ be the time after which the ball is at the top of the poles.

$$y (t) = 0. 9 m$$ $$v _ {y} (0) = v \sin 4 5 ^ {0} = 7 \sqrt {2} \frac {1}{\sqrt {2}} = 7 \frac {m}{s}.$$

Now,

$$y (t) = v _ {y} (0) t + \frac {a _ {y} t ^ {2}}{2}$$ $$0. 9 = 7 t - (9. 8) t ^ {2}$$ $$4. 9 t ^ {2} - 7 t + 0. 9 = 0$$

On solving, we get

$$t _ {1} = \frac {1}{7} s \text{ and } t _ {2} = \frac {9}{7} s.$$

Hence the ball is at A after $\frac{1}{7} s$ and at B after $\frac{9}{7} s$ .

$$O P = v _ {x} t _ {1} = 7 \frac {1}{7} = 1 m$$ $$O Q = v _ {x} t _ {2} = 7 \frac {9}{7} = 9 m$$ $$P Q = 9 - 1 = 8 m$$

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