A body is projected at time $t = 0$ from a certain point on a planet's surface with a certain velocity at a certain angle with the planet's surface (assumed horizontal). The horizontal and vertical displacement $x$ and $y$ (in metre ) respectively vary with time $t$ in second as, $x = (10 \sqrt{3}) t$ and $y = 10 t - t^{2} .$ The maximum height attained by the body is

200 m

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100 m

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50 m

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25 m

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Solution

The correct option is C $25 m$
Differentiate y to get, $v_{y} = 10 - 2 t$ . Thus maximum height means $v_{y} = 0$ , or, $t = 5 s$ .
Thus, for $t = 5$ , $y = 10 \times 5 - 5^{2} = 25 m$

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Similar questions

Q. A body is projected at time (t = 0) from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meter) respectively vary with time t in seconds as,

x = 10 \sqrt{3} t

and

y = 10 t - t^{2}

. Then, the maximum height attained by the body is :

Q. 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

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 vary with time t in seconds v as :

X = 10*3^1/2 t

Y = 10t-t^2

The maximum height by ball is?.

A ball is projected from a certain point on the ground at an angle with the horizontal surface. The horizontal and vertical displacement x and y(in metres) vary with time (in seconds) as
$x = 10 \sqrt{3} t$
$y = 10 t - 5 t^{2}$
What is the magnitude and direction of the velocity of projection?

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The correct option is C 25 mDifferentiate y to get, vy=10−2t. Thus maximum height means vy=0, or, t=5s.Thus, for t=5, y=10×5−52=25m

The correct option is C $25 m$
Differentiate y to get, $v_{y} = 10 - 2 t$ . Thus maximum height means $v_{y} = 0$ , or, $t = 5 s$ .
Thus, for $t = 5$ , $y = 10 \times 5 - 5^{2} = 25 m$