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A projectile is thrown from a point O ....

A projectile is thrown from a point O on the ground at an angle 45° from the vertical and with a speed 5√2 m/s. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, 0.5 s after the splitting. The other part, t seconds after the splitting, falls to the ground at a distance x meters from the point O. The acceleration due to gravity $g = 10 m/s^2$.

Q.1 The value of t is ___ .
Q.2 The value of x is ___ .

Solution


The part that falls vertically down to the ground has initial speed = 0. So, it falls freely taking 0.5 s to reach the ground.

The other part has horizontal velocity $2ucos45^\circ$ but even it has 0 vertical velocity. 0 vertical velocity also means vertical free fall. So, the time should be the same as for the other part. Hence the answer to Q.1 is t = 0.5 s.

Now, $x = \frac{1}{2}\frac{{{u^2}\sin (2 \times 45^\circ )}}{g} + 2u\cos 45^\circ  \times t$

$ = \frac{1}{2}\frac{{{{(5\sqrt 2 )}^2}\sin 90^\circ }}{{10}} + 2 \times 5\sqrt 2  \times \frac{1}{{\sqrt 2 }} \times 0.5$

$ = 2.5 + 5 = 7.5$

So, the answer to Q.2 is 7.5 m.

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