(C) The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.Substituting the values: $H = \frac{(25)^2 \cdot (\sin 45^{\circ})^2}{2 \time

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$h_{max} = 15.625\, m, T = 1.77\, s$

(C) The formula for maximum height is $H = \frac{u^2 \sin^2 \theta}{2g}$.Substituting the values: $H = \frac{(25)^2 \cdot (\sin 45^{\circ})^2}{2 \times 10} = \frac{625 \times 0.5}{20} = \frac{312.5}{20} = 15.625\, m$.The formula for time taken to reach the highest point is $T = \frac{u \sin \theta}{g}$.Substituting the values: $T = \frac{25 \times \sin 45^{\circ}}{10} = \frac{25 \times 0.707}{10} = 2.5 \times 0.707 = 1.7675\, s \approx 1.77\, s$.

Class 11 Physics 3-2.Motion in Plane Horizontal Projectile Motion

Medium

$A$ 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 the highest point during motion? (Take $g = 10\, ms^{-2}$)

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A
$h_{max} = 10\, m, T = 2.5\, s$
B
$h_{max} = 15.625\, m, T = 3.54\, s$
C
$h_{max} = 15.625\, m, T = 1.77\, s$
D
$h_{max} = 3.54\, m, T = 0.125\, s$

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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 the highest point during motion? (Take $g = 10\, ms^{-2}$)

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$A$ projectile is thrown at an angle of $\pi/3$ with the horizontal,reaching a maximum height $Y$. If another projectile is thrown with the same initial velocity at an angle of $\pi/6$,what will be the maximum height attained?

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