A car travelling at 20 m/s due north along the highway makes a right turn on to a side road that heads due east. It takes 50 s f
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Answer:
d = 19.796m
Explanation:
Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.
So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.
Next, you look for a kinematic equation that has these parameters and the one you should choose is:
Now by substituting values in, we get
d = 19.796m
Answer:
29,7 m
Explanation:
We need to devide the problem in two parts:
A) Energy
B) MRUV
<u>Energy:</u>
Since no friction between pint (1) and (2), then the energy conservatets:
Energy = constant ----> Ek(cinética) + Ep(potencial) = constant
Ek1 + Ep1 = Ek2 + Ep2
Ek1 = 0 ; because V1 is zero (the ball is "dropped")
Ep1 = m*g*H1
Ep2= m*g*H2
Then:
Ek2 = m*g*(H1-H2)
By definition of cinetic energy:
m*(V2)²/2 = m*g*(H1-H2) --->
Replaced values: V2 = 14,0 m/s
<u>MRUV:</u>
The decomposition of the velocity (V2), gives a for the horizontal component:
V2x = V2*cos(α)
Then the traveled distance is:
X = V2*cos(α)*t.... but what time?
The time what takes the ball hit the ground.
Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²
In the vertical axis:
Y3 = 0 ; Y2 = H2 = 2 m
Reeplacing:
-2 = 14*t + (1/2)*(-9,81)*t²
solving the ecuation, the only positive solution is:
t = 2,99 sec ≈ 3 sec
Then, for the distance:
X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m
