Answer:
28.2 m/s
Explanation:
The range of a projectile launched from the ground is given by:
[tex]d=\frac{v^2}{g}sin 2\theta[/tex]
where
v is the initial speed
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\theta[/tex] is the angle at which the projectile is thrown
In this problem we have
d = 81.1 m is the range
[tex]\theta=45^{\circ}[/tex] is the angle
Solving for v, we find the speed of the projectile:
[tex]v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s[/tex]