JEE MAIN - Physics (2021 - 26th August Morning Shift - No. 23)
Two spherical balls having equal masses with radius of 5 cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35 m/s, then these balls collide at a height of ............... m. (Take g = 10 m/s2)
Answer
50
Explanation
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When both balls will collide
y1 = y2
$$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$$
$$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$$
0 = 150 $$-$$ 30 t
t = 5 sec
$$\therefore$$ Height at which both balls will collied
$$h = 35t - {1 \over 2} \times 10 \times {t^2}$$
$$ = 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$$
h = 50 m
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