Respuesta :
That arrow goes 30.59 m high, it travels 211.96 m horizontally and taken 4.99 seconds to land.
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = -u sin θ
Acceleration = -g
Substituting
v = u + at
-u sin θ = u sin θ - g t
[tex]t=\frac{2usin\theta }{g}[/tex]
This is the time of flight.
Consider the vertical motion of ball till maximum height,
We have equation of motion v² = u² + 2as
Initial velocity, u = u sin θ
Acceleration, a = -g
Final velocity, v = 0 m/s
Substituting
v² = u² + 2as
0² = u²sin² θ + 2 x -g x H
[tex]H=\frac{u^2sin^2\theta }{2g}[/tex]
This is the maximum height reached,
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time, [tex]t=\frac{usin\theta }{g}[/tex]
Substituting
s = ut + 0.5 at²
[tex]s=ucos\theta \times \frac{2usin\theta }{g}+0.5\times 0\times (\frac{2usin\theta }{g})^2\\\\s=\frac{2u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{g}[/tex]
This is the range.
Here u = 49 m/s and θ = 30 degrees
Substituting
[tex]t=\frac{2usin\theta }{g}=\frac{2\times 49\times sin30 }{9.81}=4.99s\\\\H=\frac{u^2sin^2\theta }{2g}=\frac{49^2\times sin^230 }{2\times 9.81}=30.59m\\\\R=\frac{u^2sin2\theta}{g}=\frac{49^2\times sin(2\times 30)}{9.81}=211.96m[/tex]
That arrow goes 30.59 m high, it travels 211.96 m horizontally and taken 4.99 seconds to land.
