Question

a body is projected at an angle of 30 degree with a velocity of 9.8 metre per second when does it reach the maximum height​

Answer 1

Answer:

The time taken to reach the maximum height is 0.5 sec

Explanation:

Given :

A body is projected at an angle of 30° with a velocity of 9.8 m/s

To find :

the time taken to reach the maximum height

Solution :

  $\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,-1){\vector(0,1){5}}\put(-1,0){\vector(1,0){8}}\qbezier(0,0)(3,5)(6,-0)\multiput(3,0)(0,0.38){7}{\qbezier(0,0)(0,0)(0,0.2)}\put(3.4,1){\bf H}\put(3,-0.7){\bf R}\put(2.8,-0.6){\vector(-1,0){2.7}}\put(3.5,-0.6){\vector(1,0){2.6}}\put(0,0){\vector(1,2){1}}\put(1.1,2.3){\bf\large u}\end{picture}$

The time taken to reach the maximum height is given by,

   $\longrightarrow \underline{\boxed{ \sf t=\dfrac{u \sin \theta}{g}}}$

where

u denotes the velocity with which the body is projected

θ denotes the angle of projection

g denotes the acceleration due to gravity

• θ = 30°

• u = 9.8 m/s

• g = 9.8 m/s²

Substitute the values,

 $\longrightarrow \sf t=\dfrac{9.8 \times \sin 30^{\circ}}{9.8} \\\\ \longrightarrow \sf t=\dfrac{1}{2} \\\\ \longrightarrow t=0.5 \ sec$

Therefore, the time taken to reach the maximum height is 0.5 sec

____________________

Know more :

For an object projected at an angle θ with the horizontal,

• The maximum height is given by,

    $\bf H=\dfrac{u^2 \sin^2 \theta}{2g}$

• The range is given by,

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