Question

a body is thrown with velocity of 40 m per second in a direction making an angle of 30 degree with the horizontal calculate the horizontal range and maximum height and time taken to reach the maximum height

Answer 1

Your answer is shown below in fig. Please mark an answer as brain list.

Answer 2

Given :

Initial velocity , u = 40 m .

Angle it make with the horizontal , $\theta=30^o$ .

To Find :

The horizontal range, maximum height and time taken to reach the maximum height .

Solution :

Horizontal range is given by :

$R=\dfrac{u^2sin2\theta}{g}\\\\R=\dfrac{40^2\times sin (2\times 30)^o}{10}\\\\R=\dfrac{40^2\times \sqrt{3}}{2\times 10}\ m\\\\R=138.56 \ m$

Maximum height is given by :

$H=\dfrac{u^2sin^2\theta}{2g}\\\\H=\dfrac{40^2\times sin 30^o}{2\times 10}\\\\H=\dfrac{40^2}{2\times 2\times 10}\ m\\\\H=40 \ m$

Time taken to reach maximum height is :

$T=\dfrac{usin\theta}{g}\\\\T=\dfrac{40\times sin30^o}{10}\\\\T=\dfrac{40}{2\times 10}\\\\T=2\ s$

Hence , this is the required solution .