Question

In an I.P.L. "Chris Gayle" of R.C.B hit a six which lands on the roof of Stadium situated at a distance of 60m from gayle. If time of flight of ball is 4 sec and ball hits the stadium roof at an angle of 45° with horizontal in downward direction then.

450

C.G

Initial speed of the ball when Gayle hit the ball was:

(1)

(2)

√150 m/s

√1250 m/s

√850 m/s

✓1050 m/s

60m

Answer 1

Answer:

Benjamin Franklin showed his determination to reach Philadelphia despite his difficult and challenging journey towards this place he did not hesitate until he reached his destination even it caused sacrifice on his part.

Explanation:

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Answer 2

Answer:

The option (3) $\sqrt{850} \frac{m}{s}$ is correct.

Concept:

What is trajectory in projectile motion?

A moving object's trajectory, also known as its flight path, is the path it takes as it is pulled by gravity. Typically, the phrase is used in reference to missiles or satellites or where the regularly repeating trajectory is called an orbit). A parabola is a decent approximation of the trajectory form when an object is thrown short distances.

Given:

Distance of roof of stadium from Gayle, s = 60m

Time of flight, t = 4 sec

The stadium roof  with horizontal in downward direction, $\theta$ = $45^{\circ}$

Find:

We have to take out the initial speed of the ball when Gayle hit the ball.

Solution:

First we will take the velocity along x - direction.

$s = ut + \frac{1}{2} a t^{2} \\60 = u_{x} 4 + \frac{1}{2} \times (0) \times 4^{2}$

$u_{x} = 15\frac{m}{s}$.

Now, take the velocity along y - direction.

As we can see the final moment angle was $45^{\circ}$.

$v_{x} = v_{y} = u_{x} \\v_{y} = 15\\v = u + at$

$-15 = u_{y} - 10 \times 4 \\ u_{y} = -15 + 40\\ u_{y} = 25 \frac{m}{s}$

So, final initial speed,

$u = \sqrt{u_{x} ^{2}+u_{y} ^{2} } \\u = \sqrt{15^{2}+25^{2} }\\ u = \sqrt{225 + 625} \\u = \sqrt{850} \frac{m}{s}$

Hence, the final initial speed of the ball when Gayle hit the ball was $\sqrt{850} \frac{m}{s}$.

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