Question

A ball is projected from ground in such a way that after 10 seconds of projection it lands on ground 500 m
away from the point of projection. Find out
6) Velocity of ball after 5 seconds
(ii) angle of projection
(ii)velocity of projection​

Answer 1

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Answer 2

Answer:

(I). The Velocity of ball after 5 seconds is 50 m/s.

(II). The angle of projection is 44.42°.

(III). The velocity of projection​ is 70.00 m/s.

Explanation:

Given that,

Time t = 10 sec

Distance = 500 m

We need to calculate the vertical velocity

Using formula of time of flight

$t=\dfrac{2u\sin\theta}{g}$

Put the value into the formula

$u\sin\theta=\dfrac{10\times9.8}{2}$

$u\sin\theta=49\ m/s$

(I). We need to calculate the Velocity of ball after 5 seconds

After 5 second the vertical velocity will be zero.

So, The horizontal velocity is

$R=\dfrac{u^2\sin2\theta}{g}$

Put the value into the formula

$500=\dfrac{2u\timesu\sin\theta\times\cos\theta}{9.8}$

$2u\timesu\sin\theta\times\cos\theta=500\times9.8$

$2u\cos\theta\times49=4900$

$u\cos\theta=\dfrac{4900}{2\times49}$

$u\cos\theta=50\ m/s$

(II). We need to calculate the angle of projection

Using horizontal and vertical velocity

$u_{y}=u\sin\theta$...(I)

$u_{x}=u\cos\theta$....(II)

Divided equation (I) by equation (II)

Put the value into the formula

$\tan\theta=\dfrac{49}{50}$

$\theta=\tan^{-1}\dfrac{49}{50}$

$\theta= 44.42^{\circ}$

(III). We need to calculate the velocity of projection

Using formula of range

$R=\dfrac{u^2\sin2\theta}{g}$

Put the value into the formula

$500=\dfrac{u^2\sin2\times44.42}{9.8}$

$u^2=\dfrac{500\times9.8}{\sin2\times44.42}$

$u^2=\sqrt{4901.00}$

$u=70.00\ m/s$

Hence, (I). The Velocity of ball after 5 seconds is 50 m/s.

(II). The angle of projection is 44.42°.

(III). The velocity of projection​ is 70.00 m/s.