Aball is thrown vertically upward from the ground with an initial velocity of 60 ft/sec. how high will the ball go?

The ball will go to maximum height of 56.25 meters.

Step-by-step explanation:

It is given that, a ball is thrown vertically upward from the ground. Initial velocity of the ball, v = 60 ft/sec

Initial height is 0 as the ball is thrown from ground. The equation of projectile is given by :

h = 0

..............(1)

We need to find the maximum height reached by the ball. For maximum height,

t = 1.875 seconds

Put the values of t in equation (1). So,

h(t) = 56.25 meters

So, the ball will go to maximum height of 56.25 meters. Hence, this is teh required solution.

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

t hope this

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