Question

The horizontal and vertical components
of velocity of a projectile Is after it is
projected are 10.2 m/sand 32m/s
respectively. calculate the maximum
height attained by the projectile​

Answer 1

Solution :

As per the given data ,

• u cos Ф = 10.2 m/s

• u sin Ф  = 32 m/s
• Acceleration due to gravity = 9.8 m/s²

Maximum height attained by the projectile ,

$\dag \boxed{\mathtt{H =\dfrac{ u^2 sin^2 \phi}{2g}}}$

here ,

• H = maximum height
• u = initial velocity
• Ф = angle of projection
• g = acceleration due to gravity

On rearranging ,

➜ H = u sinФ x u sinФ / 2 x g

Now let us substitute the given values in the above equation ,

➜ H = 32 x 32 / 2 x 9.8

➜ H = 1024 / 19.6

➜ H = 52.24 m

The maximum height attained by the projectile is 52.4 m.