Question-195874

Question Number 195874 by Humble last updated on 12/Aug/23

Answered by mr W last updated on 12/Aug/23

x = v_{0} \sin θ t + \frac{g \cos α}{2} t^{2}

z = v_{0} \cos θ t - \frac{g \sin α}{2} t^{2}

a)

z_{m a x} = \frac{{(v_{0} \cos θ)}^{2}}{2 g \sin α} = \frac{{(4 \times \frac{\sqrt{3}}{2})}^{2}}{2 \times 9.8 \times \sin 45 °} = 0.87 m

b)

z = v_{0} \cos θ t - \frac{g \sin α}{2} t^{2} = 0

\Rightarrow t = \frac{2 v_{0} \cos θ}{g \sin α}

x_{m a x} = (v_{0} \sin θ + \frac{g \cos α}{2} \times \frac{2 v_{0} \cos θ}{g \sin α}) \frac{2 v_{0} \cos θ}{g \sin α}

= (\sin θ + \frac{\cos θ}{\tan α}) \frac{2 v_{0}^{2} \cos θ}{g \sin α}

= (\frac{1}{2} + \frac{\sqrt{3}}{2}) \frac{2 \times 4^{2} \times \sqrt{3} \times \sqrt{2}}{2 \times 9.8}

= 5.46 m

Commented by Humble last updated on 12/Aug/23

Great!

Thanks sir .

Facebook Tweet Pin