Aball is launched at an angle of 39.8 degrees up from the horizontal, with a muzzle velocity of 6.6 meters per second, from a launch point which is 1 meters above the floor. how high will the ball be above the floor (in meters), when it is a horizontal distance of 2.7 meters away? use 9.82 meters per second for "g".

1.85 m

Explanation:

The horizontal velocity of the ball is

The horizontal distance travelled is

d = 2.7 m

And since the motion along the horizontal direction is a uniform motion, the time taken is

The vertical position of the ball is given by

where

h = 1 m is the initial heigth

is the initial vertical velocity

g = 9.82 m/s^2 is the acceleration due to gravity

Substituting t = 0.53 s, we find the height of the ball at this time:

To convert kilometers per hour to meters per second, perform dimensional analysis. Remember that:

1 km = 1000 m

1 hr = 3600 seconds

Using these conversion, perform dimensional analysis:  

16.2 km/ hr (1000m/1km) (1hr/60 sec) = 4.5 m/s

The analysis basically just uses the conversion factors and canceling of units. The final answer is 4.5 m/s.  

Correction: That should be *(1 hr/3600 sec). The answer is still 4.5 m/s.

Hope this helps, i did the test and this answer was right, oh and brainliest, Good luck.

i think if it were me i whould divide 300 dide 15

step-by-step explanation: