Question

A ball is projected at some angle with horizontal under gravity, such that its time of flight is 10 s and range 100 m. The maximum height attained by the ball is (g = 10 m/s2)

Answer 1

Answer:

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Answer 2

Answer:

The maximum height attended by projectile(ball) is 125 m

Explanation:

Provided that:

Time of flight (t) = 10 second

Range (R) = 100 m

Gravitational acceleration (g) = 10 m/sec^2

Let's suppose the angle of projection (θ) = α

And initial velocity is u m/s

For the Projectile Motion we know;

Time of flight;

$T = \frac{2u \sin( \alpha) }{g} \: \: ...equation(i)$

And Range of Projectile;

$R = \frac{ {u}^{2} \sin(2 \alpha) }{g} \: \: ...equation(ii)$

From equation (i) we get;

$T = \frac{2u \sin( \alpha) }{g} \\ or \: u \sin( \alpha) = \frac{gT}{2} \\ or \: u \sin( \alpha) = \frac{(10 \times 10)}{2} \\ = 50 \: m. {sec}^{ - 1}$

Let the maximum height attended by projectile is H m

So now, calculating the maximum height attended;

$H = \frac{ {u}^{2} { \sin}^{2}( \alpha) }{2g} \: \: ...equation(iii) \\ so \: H = \frac{({u \sin \alpha })^{2} }{2g}$

Putting the value of u sinα ;

$H = \frac{({u \sin \alpha })^{2} }{2g} \\ = \frac{ {50}^{2} }{(2 \times 10)} \: meter \\ = 125 \: meter$

So the maximum height attended by projectile is 125 m