Question

A cricket ball is thrown at a speed of 28 m/s in a direction 30° above the horizontal. Calculate :

i ) The maximum height
ii ) The Time taken by the ball to return to the same level.
iii ) The distance from the Thrower to the point where the ball returns to the same level .​

Answer 1

Explanation:

The velocity time graph for the given time interval has been attached. Explanation: ... Initially, the ball goes up by decreasing its velocity from a positive to zero till it reaches its maximum height and then decreases its velocity from zero to negative while going down

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Projectile Motion has been used whose case is when the Projectile is fired at an angle with the horizontal. We see that we are given the initial speed of the ball and the angle which it makes with horizontal. This is a formula based question. We already know the formulas. For doing this question, we will use three formulas where we will apply values and find the answers.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{H\;=\;\bf{\dfrac{u^{2}\:\sin^{2}\theta}{2g}}}}}$

$\\\;\boxed{\sf{\pink{T\;=\;\bf{\dfrac{2\:u\sin\theta}{g}}}}}$

$\\\;\boxed{\sf{\pink{R\;=\;\bf{\dfrac{u^{2}\:\sin\:2\theta}{g}}}}}$

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★ Solution :-

Given,

» Initial velocity of ball = u = 28 m/sec

» Acceleration due to gravity = g = 9.8 m/sec²

» Angle of projection = θ = 30°

From this we get,

» sin θ = ½

» 2θ = 2 × 30° = 60°

» sin 2θ = √3/2

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i ) The Maximum Height ::

• Let the maximum height reached by the ba be H.

Then according to the formula, we get

$\\\;\sf{\rightarrow\;\;H\;=\;\bf{\dfrac{u^{2}\:\sin^{2}\theta}{2g}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;H\;=\;\bf{\dfrac{(28)^{2}\:\times\:\sin^{2}30^{\circ}}{2\:\times\:9.8}}}$

$\\\;\sf{\Longrightarrow\;\;H\;=\;\bf{\dfrac{(28)^{2}\:\times\:\bigg(\dfrac{1}{2}\bigg)^{2}}{2\:\times\:9.8}}}$

$\\\;\sf{\Longrightarrow\;\;H\;=\;\bf{\dfrac{(784)\:\times\:\bigg(\dfrac{1}{4}\bigg)}{19.6}}}$

$\\\;\sf{\Longrightarrow\;\;H\;=\;\bf{\dfrac{784\:\times\:\dfrac{1}{4}}{19.6}}}$

$\\\;\sf{\Longrightarrow\;\;H\;=\;\bf{\dfrac{196}{19.6}}}$

$\\\;\bf{\Longrightarrow\;\;H\;=\;\bf{\green{10\;\;m}}}$

$\\\;\underline{\boxed{\tt{Maximum\;\:height\;\:reached\;\:by\;\:ball\;=\;\bf{\purple{10\;\;m}}}}}$

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ii ) The Time Taken by the ball to return the same level ::

We see that here it is asked for time taken to return at same level. This means we need to add the time for going up and time for coming down. This total time is known as Time of Flight of Projectile.

• Let the time taken by the ball be T

So let's calculate this which according to the formula is given as,

$\\\;\sf{\rightarrow\;\;T\;=\;\bf{\dfrac{2\:u\sin\theta}{g}}}$

By applying the values, we get

$\\\;\sf{\Longrightarrow\;\;T\;=\;\bf{\dfrac{2\:\times\:28\:\times\:\bigg(\dfrac{1}{2}\bigg)}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;T\;=\;\bf{\dfrac{52\:\times\:\bigg(\dfrac{1}{2}\bigg)}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;T\;=\;\bf{\dfrac{52\:\times\:\dfrac{1}{2}}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;T\;=\;\bf{\dfrac{26}{9.8}}}$

$\\\;\bf{\Longrightarrow\;\;T\;=\;\bf{\blue{2.86\;\;seconds}}}$

$\\\;\underline{\boxed{\tt{Time\;\:taken\;\:by\;\:the\;\:ball\;=\;\bf{\purple{2.86\;\;seconds}}}}}$

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iii ) The Distance from the Thrower to the point where the ball returns to the same level ::

Here we see that we have to calculate the range of the Projectile. We see that the distance from the thrower to the point where the ball returns is Range only.

• Let the range of the projectile be R

We know that,

$\\\;\sf{\rightarrow\;\;R\;=\;\bf{\dfrac{u^{2}\:\sin\:2\theta}{g}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{(28)^{2}\:\times\:\sin\:60^{\circ}}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{(28)^{2}\:\times\:\bigg(\dfrac{\sqrt{3}}{2}\bigg)}{9.8}}}$

Since, √3 = 1.73

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{(28)^{2}\:\times\:\bigg(\dfrac{1.73}{2}\bigg)}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{(784)\:\times\:\bigg(0.866\bigg)}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{784\:\times\:0.866}{9.8}}}$

$\\\;\sf{\Longrightarrow\;\;R\;=\;\bf{\dfrac{678.944}{9.8}}}$

$\\\;\bf{\Longrightarrow\;\;R\;=\;\bf{\red{69.28\;\;m}}}$

$\\\;\underline{\boxed{\tt{Required\;\:distance\;\:of\;\:thrower\;\:from\;\:ball\;=\;\bf{\purple{69.28\;\;m}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Time\;of\;Maximum\;Height\;=\;\dfrac{u\:\sin\theta}{g}}$

$\\\;\sf{\leadsto\;\;Maximum\;Horizontal\;Range\;=\;\dfrac{u^{2}}{g}}$

These were for the cases when Projectile is fired with some angle along the horizontal.

• For the Projectile which is fired parallel to Horizontal ::

$\\\;\sf{\leadsto\;\;Velocity\;=\;\sqrt{u^{2}\:+\:(g^{2}t^{2})}}$

$\\\;\sf{\leadsto\;\;Time\;of\;Flight\;=\;\sqrt{\dfrac{2h}{g}}}$

$\\\;\sf{\leadsto\;\;Horizontal\;Range\;=\;u\sqrt{\dfrac{2h}{g}}}$