Solution:
We Know that,
\[ \frac{R}{H}=\frac{4}{tan\theta}\]
\[ tan \theta = 4/3 \]
\[ H = \frac{u^{2}sin^{2}\theta}{2g} \]
Solving we get
\[ u= 5\sqrt{g/2}\]
The maximum height reached by projectile is 4 m. The horizontal range is 12m. The velocity of projection in ms–¹ is : (g is acceleration due to gravity) :
Leave a Reply