Question

find this question related to kinematics projectile motion

Answer 1

5s are answer of tis ouestion

Answer 2

Time of Flight :

$tf = \frac{2u \sin(30) }{g} = \frac{2 \times 49 \times \sin(30) }{2 \times 4.9} \\ tf = 5$

Maximum height :

$= \frac{ {u}^{2} \sin {}^{2} (30) }{2g} = \frac{ {49}^{2} \times \frac{1}{4} }{2 \times 9.8} \\ = 30.6m$

Horizontal range :

$= \frac{ {u}^{2} \sin(60) }{g} = \frac{ {49}^{2} \times \frac{ \sqrt{3} }{2} }{2 \times 4.9} \\ = 122.5 \sqrt{3}$

Hope it's helpful to you