Question

A object is thrown with initial speed 5 m s⁻¹ with an angle of projection 30γ . What is the height and range reached by the particle?

Answer 1

Final Answer : Height reached = 5/16m
Range = (5√3)/4 m

Steps:
1) Initial velocity, u = 5m/s
Angle of Projection, Theta = 30°.
acceleration due to gravity, g = 10m/s^2

We know, that

Height Reached =
$\frac{ {u}^{2} {( \sin( \theta) )}^{2} }{2g}$
=>
$\frac{ {5}^{2} \times { (\sin(30 \degree)})^{2} }{2 \times 10} = \frac{5}{16} m$


Range, =
$\frac{ {u}^{2} \sin(2 \theta) }{g} = \frac{ {5}^{2} \times \sin(60 \degree) }{10} = \frac{5 \sqrt{3} }{4} m$


Hence, We get
Height Reached as 5/16m
and Range as, (5√3)/4.