Respuesta :
Explanation:
It is given that,
Height above which the stone was thrown, h = 10 m
Initial velocity of the stone, u = 8 m/s
Angle above the horizontal, [tex]\theta=25^{\circ}[/tex]
The horizontal component of velocity is, [tex]u_x=v\ cos\theta=8\ cos(25)=7.25\ m/s[/tex]
The vertical component of velocity is, [tex]u_y=v\ sin\theta=8\ cos(25)=3.38\ m/s[/tex]
Let t is the time of flight in vertical motion. The second equation of motion is :
[tex]h=u_yt-\dfrac{1}{2}gt^2[/tex]
[tex]10=3.38t-\dfrac{1}{2}\times 9.8t^2[/tex]
t = 0.34 seconds
Let s is the range of the stone. It can be calculated as :
[tex]s=u_x\ cos\theta \times t[/tex]
[tex]s=7.25\times 0.34[/tex]
s = 2.46 meters
So, the range of the stone is 2.46 meters. Hence, this is the required solution.
