Answer:
15.77m/s
Explanation:
the information we have is:
initial velocity [tex]v_{0}=16.2m/s[/tex]
distance: [tex]d=25m[/tex]
angle: [tex]\theta=38[/tex]°
first we need to break down the velocity into its x and y components:
initial velocity in x (the velocity in x is constant):
[tex]v_{0x}=v_{0}cos\theta\\v_{0x}=(16.2m/s)cos38\\v_{0x}=12.766m/s[/tex]
and initial velocity in y (the velocity in y is not constant due to acceleration of gravity):
[tex]v_{0y}=v_{0}sin\theta\\v_{0y}=(16.2m/s)sin38\\v_{0y}=9.97m/s[/tex]
and now we find the time that the ball was in the air:
[tex]t=\frac{d}{v_{0x}} =\frac{25m}{12.766m/s}\\ t=1.96s[/tex]
and with this time, we find the y component of the volicity at time 1.96s:
[tex]v_{y}=v_{0y}-gt\\v_{y}=(9.97m/s)-(9.81m/s^2)(1.96s)\\v_{y}=-9.26m/s[/tex](negative because it points downward)
finally, to find the final velocity we use pythagoras:
[tex]v_{f}=\sqrt{v_{x}^2+v_{y}^2} =\sqrt{12.766^2+(-9.26)^2}=15.77m/s[/tex]
