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Byju's Answer

Standard XII

Physics

Graph of Variation of G

A ball is pro...

Question

A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement $x$ and $y$ varies with time $t$ in second as:
$x = 10 \sqrt{3} t$ and $y = 10 t - t^{2}$
The maximum height attained by the ball is

100 m

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75 m

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50 m

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25 m

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Solution

The correct option is C $25 m$
The graph of the y coordinate is a parabola.
The parabola is symmetric between t = 0 to t =10 (about t = 5).
Hence the maximum value would be achieved at t = 5s.
$y = 10 t - t^{2} = (10 \times 5) - (5^{2}) = 25 m$
Hence option $D$ is correct.

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A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement x and y varies with time t in second as:x=10√3t and y=10t−t2The maximum height attained by the ball is

The correct option is C 25mThe graph of the y coordinate is a parabola.The parabola is symmetric between t = 0 to t =10 (about t = 5).Hence the maximum value would be achieved at t = 5s.y=10t−t2=(10×5)−(52)=25mHence option D is correct.

A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement $x$ and $y$ varies with time $t$ in second as:
$x = 10 \sqrt{3} t$ and $y = 10 t - t^{2}$
The maximum height attained by the ball is

The correct option is C $25 m$
The graph of the y coordinate is a parabola.
The parabola is symmetric between t = 0 to t =10 (about t = 5).
Hence the maximum value would be achieved at t = 5s.
$y = 10 t - t^{2} = (10 \times 5) - (5^{2}) = 25 m$
Hence option $D$ is correct.