Respuesta :
Let x = the angle of elevation for shooting the arrow.
Assume
g = 9.8 m/s²
No wind resistance
The vertical launch velocity is 25.1 sin(x) m/s
The horizontal velocity is 25.1 cos(x) m/s
The time of flight is
24/[25.1 cos(x)] s = 0.9562 sec(x) s
Therefore
0.5*[0.9562 sec(x)]*(9.8) = 25.1 sin(x)
4.6854 = 25.1* sin(x)cos(x)
sin(2x) = 0.3733
2x = sin⁻¹ 0.3733 = 21.92 deg
x = 10.96 deg
Answer: 11 degrees (nearest integer)
Assume
g = 9.8 m/s²
No wind resistance
The vertical launch velocity is 25.1 sin(x) m/s
The horizontal velocity is 25.1 cos(x) m/s
The time of flight is
24/[25.1 cos(x)] s = 0.9562 sec(x) s
Therefore
0.5*[0.9562 sec(x)]*(9.8) = 25.1 sin(x)
4.6854 = 25.1* sin(x)cos(x)
sin(2x) = 0.3733
2x = sin⁻¹ 0.3733 = 21.92 deg
x = 10.96 deg
Answer: 11 degrees (nearest integer)
Answer : Angle is [tex]\theta=11^0[/tex]
Explanation :
It is given that,
Horizontal range, R = 24 m
Velocity, v = 25.1 m/s
We have to find the angle with which he hit the apple.
We know that horizontal range is given by the relation as :
[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]
Putting all values in above equation we get:
[tex]sin2\theta=\dfrac{R\ g}{v^2}[/tex]
[tex]sin2\theta=\dfrac{24\ m\times 9.8\ m/s^2}{(25.1\ m)^2}[/tex]
[tex]sin2\theta=0.373[/tex]
[tex]\theta=10.95^0[/tex]
or
[tex]\theta=11^0[/tex]
So, the angle must he aim it to hit the apple is [tex]\theta=11^0[/tex].