A football is kicked with a velocity of 20 m - s at an angle of 45- with the horizontal- Find the maximum height it reaches-

At the highest point, u_y = 0. Using the equation v^2_y=u^2_y 2gy, 0=(10√(2)m/s)^2 2×(10m/s^2)H Or, H = 10m. Thus, The maximum height reached is 10 m. ...

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Standard XII

Physics

Average Velocity in 2D Motion

A football is...

Question

A football is kicked with a velocity of 20 m/s at an angle of $45^{\circ}$ with the horizontal. Find the maximum height it reaches.

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Solution

At the highest point, $u_{y}$ = 0. Using the equation
$v_{y}^{2} = u_{y}^{2} - 2 g y,$
0= $(10 \sqrt{2} \frac{m}{s})^{2} - 2 \times (10 \frac{m}{s^{2}}) H$
Or, H = 10m.
Thus, The maximum height reached is 10 m.

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A football is kicked with a velocity of 20 m/s at an angle of 45∘ with the horizontal. Find the maximum height it reaches. ___

At the highest point, uy = 0. Using the equation v2y=u2y−2gy, 0=(10√2ms)2−2×(10ms2)H Or, H = 10m. Thus, The maximum height reached is 10 m.

A football is kicked with a velocity of 20 m/s at an angle of $45^{\circ}$ with the horizontal. Find the maximum height it reaches.

___

At the highest point, $u_{y}$ = 0. Using the equation
$v_{y}^{2} = u_{y}^{2} - 2 g y,$
0= $(10 \sqrt{2} \frac{m}{s})^{2} - 2 \times (10 \frac{m}{s^{2}}) H$
Or, H = 10m.
Thus, The maximum height reached is 10 m.