The driver of a car slams on the brakes, causing the car to slow down at a rate of
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Answer:
a) The maximum height reached by the projectile is 558 m.
b) The projectile was 21.3 s in the air.
Explanation:
The position and velocity of the projectile at any time "t" is given by the following vectors:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position
v0 = initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).
v = velocity vector at time t
a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:
vy = v0 · sin α + g · t
0 = 175 m/s · sin 36.7° - 9.80 m/s² · t
- 175 m/s · sin 36.7° / - 9.80 m/s² = t
t = 10.7 s
Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).
Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.
y = y0 + v0 · t · sin α + 1/2 · g · t²
y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²
y = 558 m
The maximum height reached by the projectile is 558 m
b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.
We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²
0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)
0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t
- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t
t = 21.3 s
The projectile was 21.3 s in the air.
Answer:
The kangaroo was 1.164s in the air before returning to Earth
Explanation:
For this we are going to use the equation of distance for an uniformly accelerated movement, that is:
Where:
x = Final distance
xo = Initial point
Vo = Initial velocity
a = Acceleration
t = time
We have the following values:
x = 1.66m
xo = 0m (the kangaroo starts from the floor)
Vo = 0 m/s (each jump starts from the floor and from a resting position)
a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)
t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.
Now replace the values in the equation
It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is
t = 0.582s + 0.582s
t = 1.164s
The kangaroo was 1.164s in the air before returning to Earth