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3-2.Motion in Plane
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A player kicks a football with an initial speed of $25\, {ms}^{-1}$ at an angle of $45^{\circ}$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g $=10 \,{ms}^{-2}$ )
A${h}_{\max }=10\, {m} \quad {T}=2.5\, {s}$
B${h}_{\max }=15.625\, {m} \quad {T}=3.54 \,{s}$
C${h}_{\max }=15.625\, {m} \quad {T}=1.77 \,{s}$
D${h}_{\max }=3.54 \,{m} \quad {T}=0.125 \,{s}$
(JEE MAIN-2021)
Solution
${H} =\frac{{U}^{2} \sin ^{2} \theta}{2 {g}}$
$=\frac{(25)^{2} \cdot(\sin 45)^{2}}{2 \times 10}$
$=15.625 {m}$
${T} =\frac{{U} \sin \theta}{{g}}$
$=\frac{25 \times \sin 45^{\circ}}{10}$
$=2.5 \times 0.7$
$=1.77\, {s}$
$=\frac{(25)^{2} \cdot(\sin 45)^{2}}{2 \times 10}$
$=15.625 {m}$
${T} =\frac{{U} \sin \theta}{{g}}$
$=\frac{25 \times \sin 45^{\circ}}{10}$
$=2.5 \times 0.7$
$=1.77\, {s}$
Standard 11
Physics
Similar Questions
Given that $u_x=$ horizontal component of initial velocity of a projectile, $u_y=$ vertical component of initial velocity, $R=$ horizontal range, $T=$ time of flight and $H=$ maximum height of projectile. Now match the following two columns.
| Column $I$ | Column $II$ |
| $(A)$ $u_x$ is doubled, $u_y$ is halved | $(p)$ $H$ will remain unchanged |
| $(B)$ $u_y$ is doubled $u_x$ is halved | $(q)$ $R$ will remain unchanged |
| $(C)$ $u_x$ and $u_y$ both are doubled | $(r)$ $R$ will become four times |
| $(D)$ Only $u_y$ is doubled | $(s)$ $H$ will become four times |
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