Abaseball is launched horizontally from a height of 1.8 m. the baseball travels 0.5 m before hitting the ground.
the velocity of the baseball, rounded to the nearest hundredth, is m/s.

0.82 m/s

Explanation:

This is a projectile motion, consisting of a uniform motion along the horizontal direction and a uniformly accelerated motion along the vertical direction.

We know that the baseball is launched from a height of 1.8 m. The initial vertical velocity is zero, so we can find the time taken to hit the ground by using the equation for the uniformly accelerated motion:

where h=1.8 m and g=9.81 m/s^2 is the acceleration due to gravity. Re-arranging the formula, we find t

We know that during this time, the ball moves along the horizontal direction by S=0.5 m. Since the horizontal velocity of the ball is constant, we can calculate it:

d

explanation:

i think the answer is d. but im not sure!

v=ir = 3x15 =45 v