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PART - 1 CH - 3 Motion in a Plane — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 1 CH - 3 Motion in a Plane2 Marks
Question
A ball is thrown with a velocity of $15 ms^{-1}$ at angle $45^{\circ}$ with the horizontal. What is the range of ball? What is the time of flight for the ball to return to same plane from the point of throwing?
✓

Answer

 It is given,
$\begin{aligned}
\mu & =15 m / s \\
\theta & =45^{\circ} \\
Range, R & =?
\end{aligned}$

Time of flight $T =$ ?
$g=10 m / s^2$
Range of ball $\quad R =\frac{u^2 \sin 2 \theta}{g}$
$\begin{array}{l}
=\frac{(15)^2 \sin 2 \times 45^{\circ}}{10} \\
=\frac{225 \sin 90^{\circ}}{10}=\frac{225 \times 1}{10} \\
\because \sin 90^{\circ}=1
\end{array}$
Horizontal range $R =22.5 m$
Time of flight
$\begin{aligned}
T & =\frac{2 u \sin \theta}{g} \\
& =\frac{2 \times 15 \times \sin 45^{\circ}}{10} \\
& =\frac{30 \times \frac{1}{\sqrt{2}}}{10} \\
T & =\frac{3}{\sqrt{2}} sec.
\end{aligned}$

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