Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
[tex]V=Vo+a*t[/tex] (1)
[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)
[tex]V^2=Vo^2+2*a*X[/tex] (3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)
[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]
the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.
[tex]X=200m[/tex]
[tex]V=0+25*4[/tex]
[tex]V=100m/s[/tex]
b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.
[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]
c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.
[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]
the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:
[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]
solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
