Question:

A ball is thrown at a speed of 50 ms^-1 at an angle of 60⁰ with the horizontal.Find the maximum height reached and the range of the ball (Take g= 10 ms^-2 )

Solution:

Finding the maximum height of the ball :

Initial velocity of the ball u =50 ms^-1

Angle made with the horizontal

$${\theta = 60 \ degree}$$

Maximum height of the projectile

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

$${H=\frac{u^2sin^2(60)}{2g}}$$

$${H=\frac{(50^2)(\frac{\sqrt3}{2})^2} {(2)(10)}}$$

$${H=\frac{(2500)(3)}{80}}$$

$${H=93.75 \ m}$$

Range of the ball

Range of the ball

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

$${R=\frac{(50^2)(sin120^0)}{10}}$$

$${R=\frac{(2500)(\frac{\sqrt3}{2})} {10}}$$

$${R=(125)(1.732)}$$

$${R=216.5 \ m}$$

Range of the ball is 216.5 m