student throwing a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground

Question

i really need help
please!

Answer 1

You didn't actually ask a question.
Usually these statements are followed by,
"How far from the base of the cliff did the baseball land?"
and sometimes "how long was the ball in the air?"
and maybe " what was the speed of the ball just before impact?.

If air resistance is ignored, the horizontal speed remains constant through the flight of the ball, so
x(t)=25*t

The vertical speed is accelerated from zero by gravity according to
vy(t)=-g*t
and the vertical displacement is
y(t)=y0-.5*g*t^2
in this case
y(t)=45-.5*g*t^2

g is the acceleration due to gravity 9.81 m/s^2

to find the range of the ball, solve for y(t)=0
0=45-.5*g*t^2

There are two answers since t is squared. We only care about the positive t in this case

t=3.03 seconds
plug that into x(t)
x(3.03)=25*3.03
the range is 75.72 m from the base of the cliff horizontally

the speed of the ball just before impact has two components (it's a vector)
x=25 m/s
and
vy(3.03)=-9.81*3.03
-29.71 m/s

The magnitude of the vector is
=SQRT(25^2+29.71^2)
38.83 m/s
and the angle is
=atan(29.71/25)
or 50 degrees below the horizontal

j

Answer 2

if the preliminary action is horizontal, then we are in a position to deal with the vertical action of the item as though it replaced into dropped from relax; the time of flight is then the time it takes an merchandise to fall 45m as quickly as dropped from relax we use height = a million/2 g t^2 or t = Sqrt[2 h/g] = Sqrt[2 x 45m/9.8m/s/s] = 3.0s this suggests the item would be interior the air for 3s; at a horizontal velocity of 25m/s, the item will shuttle a distance of 3s x 25m/s = 75m in the previous hitting the floor ans is a million