First slide

Projectile motion

Question

A particle A is projected from the ground with an initial velocity of 10 m/s at an angle of 60^o with horizontal. From what height h should an another particle B be projected horizontally with velocity 5 m/s so that both the particles collide in ground at point C if both are projected simultaneously $(g = 10 m / s^{2})$

Difficult

A

10 m
B

30 m
C

15 m
D

25 m

Solution

Horizontal component of velocity of A is, 10 cos 60^o or 5 m/s which is equal to the velocity of B in horizontal direction. They will collide at C if time of flight of the particles are equal or
$t_{A} = t_{B}; \frac{2 u \sin θ}{g} = \sqrt{\frac{2 h}{g}} (h = \frac{1}{2} g t_{B}^{2})$
$or h = \frac{2 u^{2} \sin^{2} θ}{g}; = \frac{2 (10)^{2} {(\frac{\sqrt{3}}{2})}^{2}}{10} = 15 m$

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