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2.Motion in Straight Line
hard
Two spherical balls having equal masses with radius of $5\, {cm}$ each are thrown upwards along the same vertical direction at an interval of $3\, {s}$ with the same initial velocity of $35 \,{m} / {s}$, then these balls collide at a height of $\ldots \ldots \ldots . {m}$. (Take ${g}=10 \,{m} / {s}^{2}$ )
A$80$
B$10$
C$60$
D$50$
(JEE MAIN-2021)
Solution
When both balls will collied${y}_{1}={y}_{2}$
$35 {t}-\frac{1}{2} \times 10 \times {t}^{2}=35({t}-3)-\frac{1}{2} \times 10 \times({t}-3)^{2}$
$35 {t}-\frac{1}{2} \times 10 \times {t}^{2}=35 {t}-105-\frac{1}{2} \times 10 \times {t}^{2}$
$\quad-\frac{1}{2} \times 10 \times 3^{2}+\frac{1}{2} \times 10 \times 6 {t}$
$0=150-30 {t}$
${t}=5 {sec}$
$\therefore$ Height at which both balls will collied
${h}=35 {t}-\frac{1}{2} \times 10 \times {t}^{2}$
$=35 \times 5-\frac{1}{2} \times 10 \times 5^{2}$
${h}=50 {m}$
Standard 11
Physics
