Answer:
x=502.837m
v=61m/S
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for the part a we can use the ecuation number 2
a=3.7m/s^2
Vf=61m/s
Vo=0m/s
[tex]\frac{Vf^{2}-Vo^2}{2.a} =X \\[/tex]
[tex]\frac{61^{2}-0^2}{2(3.7)} =X\\x=502.84m\\[/tex]
for the part b
the problem indicates that the speed with which it touches the ground is 61m / s
