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3-2.Motion in Plane
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A ball is projected upwards from the top of a tower with a velocity of $50\, ms^{-1}$ making an angle of $30^o$ with the horizontal. The height of the tower is $70\, m$. After how many seconds from the instant of throwing will the ball reach the ground ?.......$s$
A$2 $
B$5 $
C$7$
D$9$
Solution
The vertically upward component of the velocity of projection $=50$ sin $30^{\circ} \mathrm{m} / \mathrm{s}=25 \mathrm{ms}^{-1}$
If $t$ is the time taken to reach the round, we have
$\mathrm{s}=\mathrm{u}_{0} \mathrm{t}+\frac{1}{2} \mathrm{gt}^{2}$
or $\quad 70=-25 \times t+\frac{1}{2} \times 10 \times t^{2}$
or $5 \mathrm{t}^{2}-25 \mathrm{t}-70=0$ so $\mathrm{t}=-2 \mathrm{s}$ or $\mathrm{t}=7 \mathrm{s}$
If $t$ is the time taken to reach the round, we have
$\mathrm{s}=\mathrm{u}_{0} \mathrm{t}+\frac{1}{2} \mathrm{gt}^{2}$
or $\quad 70=-25 \times t+\frac{1}{2} \times 10 \times t^{2}$
or $5 \mathrm{t}^{2}-25 \mathrm{t}-70=0$ so $\mathrm{t}=-2 \mathrm{s}$ or $\mathrm{t}=7 \mathrm{s}$
Standard 11
Physics
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