A police car is traveling at a speed of 50 meters per second when it suddenly accelerates at a rate of 5 m/s2. How fast will the
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Answer:
a) Please, see the attached figure.
b) The horizontal component of the initial velocity is 8.5 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) The clown will land safely on the mattress, 10.2 m from the cannon. If we include air resistance in the calculation, he will surely not reach the mattress because, without air resistance, he lands just 20 cm from the closest edge of the mattress.
Explanation:
Hi!
a) Please, see the attached figure.
b) As shown in the figure, the initial velocity vector is the following:
v0 = (v0x, v0y)
Using trigonomety of right triangles:
cos angle = adjacent side / hypotenuse
In this case:
Adjacent side = v0x
hypotenuse = v0
(see figure)
Then:
cos 35° = v0x / v0
v0 · cos 35° = v0x
v0x = 10.4 m/s · cos 35°
v0x = 8.5 m/s
The horizontal component of the initial velocity is 8.5 m/s
We proceed in the same way to find the vertical component:
sin angle = opposite side / hypotenuse
sin 35° = v0y / v0 (see figure to notice that opposite side = v0y)
v0 · sin 35° = v0y
10.4 m/s · sin 35° = v0y
v0y = 6.0 m/s
The vertical component of the initial velocity is 6.0 m/s
c) The equation of the position vector of the clown at time t is the following:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
We have to find the time at which the vertical component of the position vector is 1 m. Let´s place the origin of the system of reference at the point where the cannon is located on the ground so that x0 = 0 and y0 = 1.0 m.
Using the equation of the vertical component of the position:
y = y0 + v0 · t · sin α + 1/2 · g · t²
1.0 m = 1.0 m + 10.4 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²
0 = 10.4 m/s · t · sin 35° - 4.9 m/s² · t²
0 = t (10.4 m/s · sin 35° - 4.9 m/s² · t) (t = 0, when t = 0 the clown is 1.0 m above the ground, just leaving the cannon).
0 = 10.4 m/s · sin 35° - 4.9 m/s² · t
-10.4 m/s · sin 35° / -4.9 m/s² = t
t = 1.2 s
The clown will return to a height of 1.0 m after 1.2 s of the launch.
d) Now, let´s calculate the horizontal traveled distance after 1.2 s using the equation of the horizontal component of the position vector:
x = x0 + v0 · t · cos α (x0 = 0)
x = 10.4 m/s · 1.2 s · cos 35°
x = 10.2 m
Since the mattress is located at 10 m from the cannon and it is 2.0 m long, the clown will land safely on the mattress. However, the clown almost miss the mattress (he lands just 20 cm from the closest edge), so, if we include air resistance in the calculation, he will surely not reach the mattress.
