Question

A football player hits a ball of an angle of 60 degree with an initial velocity of 15 metre per second .Assume that the ball travella in veRticle plane calculate 1_height 2_range 3_time of flight...​

Answer 1

Answer:-

H = 8.43 m

R = 19.48 m

T = 2.59 s

Given :-

$\theta = 60^{\circ} \\ u = 15 m/s \\g = 10 m/s^2$

To find :-

The height, range and time of flight.

Solution:-

Since, the body is projected with an angle of 60° from horizontal.

The maximum height covered by the football is :-

$\huge \boxed{H = \dfrac{u^2Sin^2 \theta }{2g}}$

• Put the given values,

→$H = \dfrac{(15)^2 \times (Sin60^{\circ})^2}{2 \times 10}$

→$H = \dfrac{225 \times (\dfrac{\sqrt{3}}{2})^2}{20}$

→$H = \dfrac{225 \times \dfrac{3}{4}}{20}$

→$H = \dfrac{\dfrac{675}{4}}{20}$

→$H = \dfrac{675}{80}$

→$H = 8.43m$

Now,

Range of the football will be :-

$\huge \boxed{R = \dfrac{u^2Sin\theta}{g}}$

→$R = \dfrac{(15)^2 Sin60^{\circ}}{10}$

→$R = \dfrac{225\times \dfrac{\sqrt{3}}{2}}{10}$

→$R = \dfrac{225\sqrt{3}}{20}$

→$R = 19.48 m$

hence,

Range of football is 19.48 m

Time of flight of football :-

$\huge \boxed{T = \dfrac{2uSin\theta}{g}}$

→$T = \dfrac{2 \times 15 Sin60^{\circ}}{10}$

→$T = \dfrac{30 \times \dfrac{\sqrt{3}}{2}}{10}$

→$T = \dfrac{15 \sqrt{3}}{10}$

→$T = 2.59 s$

hence,

Time of flight will be 2.59s.

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

Height = 8.43 m

Range = 19.48 m

Time of the flight = 2.59 s

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given that :

$\longrightarrow\sf \theta = 60^{\circ}$

$\longrightarrow\sf u = 15m/s.$

$\longrightarrow\sf g = 10m/s^2$

To find :

Height, range and time of flight.  

Solution :

The body is projected by an angle of 60° from horizontal.

We know that :

The maximum height covered by the football :

★ $\boxed{\sf{H=\dfrac{u^2\;Sin^2\theta}{2g}}}$

Plug the given values in equation :

$\longrightarrow\sf H \dfrac{(15)^2 \times (Sin60^{\circ})^2}{2 \times 10}}$

$\longrightarrow\sf H = \dfrac{225 \times (\dfrac{\sqrt{3}}{2})^2}{20}$

$\longrightarrow\sf H = \dfrac{225 \times (\dfrac{\sqrt{3}}{2})^2}{20}$

$\longrightarrow\sf H = \dfrac{225 \times \dfrac{3}{4}}{20}$

$\longrightarrow\sf H = \dfrac{\dfrac{675}{4}}{20}$

$\longrightarrow\sf H = \dfrac{675}{80}$

$\longrightarrow\sf H = 8.43m$

Hence,

It implies that Height of the flight is 8.43.

$\rule{300}{1.5}$

We know that,

Range of the football will be :

★ $\boxed{\sf R = \dfrac{u^2Sin\theta}{g}}$

Put the given values :

$\longrightarrow\sf R = \dfrac{(15)^2 Sin60^{\circ}}{10}$

$\longrightarrow\sf R =\dfrac{225\times \dfrac{\sqrt{3}}{2}}{10}$

$\longrightarrow\sf R = \dfrac{225\sqrt{3}}{20}$

$\longrightarrow\sf R = 19.48m$

Hence,

It implies that the Range of football is 19.48 m.

$\rule{300}{1.5}$

We know that,

Time of flight of football :

★ $\boxed{T = \dfrac{2uSin\theta}{g}}$

Put the given values :

$\longrightarrow\sf T = \dfrac{2 \times 15 Sin60^{\circ}}{10}$

$\longrightarrow\sf T = \dfrac{30 \times \dfrac{\sqrt{3}}{2}}{10}$

$\longrightarrow\sf T = \dfrac{15 \sqrt{3}}{10}$

$\longrightarrow\sf T = 2.59s$

Hence,

It implies that the Time of flight will be 2.59s.

$\rule{300}{1.5}$