Answer:
28.2 m/s
Step-by-step explanation:
The range of a projectile launched from the ground is given by:

where
v is the initial speed
g = 9.8 m/s^2 is the acceleration of gravity
is the angle at which the projectile is thrown
In this problem we have
d = 81.1 m is the range
is the angle
Solving for v, we find the speed of the projectile:
