Question

s
A ball is thrown with speed 20" at an angle 30° with
vertical. Its height above the point of projection when it
is at horizontal distance of 5 m from the thrower is
approximately (g = 10 ms-2)
7.41 m
8.41 m
6.14 m
5.14 m​

Answer 1

Given :

Initial velocity = 20 m/s

Angle with vertical axis = 30°

To Find :

Height of ball from the ground when it is at horizontal distance of 5m from the point of projection.

Solution :

■ A body is said to be projectile if it is projected into space with some initial velocity and then it continues to move in a vertical plane such that its horizontal acceleration is zero and vertical downward acceleration is equal to g.

★ Equation of trajectory is given by;

$\dag\:\underline{\boxed{\bf{\orange{y=x\ tan\theta-\dfrac{gx^2}{2u^2cos^2\theta}}}}}$

• x denoted horizontal distance
• y denotes vertical distance
• g denotes acc. due to gravity
• u denotes initial velocity

Angle of projection = (90 - 30) = 60°

$\mapsto\sf\:y=x\ tan\theta-\dfrac{gx^2}{2u^2cos^2\theta}$

$\mapsto\sf\:y=5\ tan60^{\circ}-\dfrac{(10)(20)^2}{2(5)^2cos^260^{\circ}}$

$\mapsto\sf\:y=5(\sqrt3)-\dfrac{250\times 4}{800}$

$\mapsto\sf\:y=8.65-\dfrac{1000}{800}$

$\mapsto\sf\:y=8.65-1.25$

$:\implies\:\underline{\boxed{\bf{\pink{y=7.4\ m}}}}$

∴ (A) is the correct answer!