A projectile of mass 0.99 kg is shot from a
cannon. The end of the cannon’s barrel is
at height 6.6 m, as shown in the figure. The
initial velocity of the projectile is 9.2 m/s.
The projectile rises to a maximum height of
∆y above the end of the cannon’s barrel and
strikes the ground a horizontal distance ∆x
past the end of the cannon’s barrel.
Find the time it takes for the projectile to
reach its maximum height.
The acceleration
of gravity is 9.8 m/s
Answer in units of s
the angle given is 47◦
b)How long does it take the projectile to hit the
ground?
Answer in units of s
c)Find the range ∆x of the projectile.
Answer in units of m
Vo = 9.2m/s[47o]
Xo = 9.2*Cos47 = 6.27 m/s.
Yo = 9.2*sin47 = 6.73 m/s.
a. Y = Yo + g*Tr = 0 @ max ht.
Tr = -Yo/g = -6.73/-9.8 = 0.687 s. =
Rise time or time to reach max ht.
b. h max = ho + Yo*Tr + 0.5g*Tr^2 =
6.6 + 6.73*0.687 - 4.9*0.687^2 = 8.91 m.
Above gnd.
c. h = 0.5g*Tf^2 = 8.91 m.
4.9*Tf^2 = 8.91
Tf^2 = 1.82
TF = 1.35 s. = Fall time.
Range = Xo*(Tr+Tf)=6.27 * (0.687+1.35) =
12.77 m.
a) Well, let's solve this like a clown firing rubber chickens out of a cannon! The projectile goes up, reaches its maximum height, and then falls back down. To find the time it takes to reach the maximum height, we can use the formula for the time of flight. Since the initial vertical velocity is 9.2 m/s and the acceleration due to gravity is 9.8 m/s^2, the time to reach the maximum height can be found using the formula: time = (final velocity - initial velocity) / acceleration. Plugging in the values, we get: time = (0 - 9.2) / (-9.8). If my calculations are correct, the answer is approximately 0.938 seconds.
b) Now, let's move on to the time it takes for the projectile to hit the ground. Since we know the initial vertical velocity, we can use it to find the time of flight to reach the ground. The time it takes to hit the ground can be found using the formula: time = (final velocity - initial velocity) / acceleration. This time, the final velocity is going to be 0 m/s since the projectile hits the ground, so the formula becomes: time = (0 - 9.2) / (-9.8). If my calculations haven't gone bananas, the answer is approximately 0.939 seconds.
c) Finally, let's calculate the range of the projectile, ∆x. The horizontal range can be found using the formula: ∆x = initial velocity * time. Plugging in the values, we get: ∆x = 9.2 * 0.939. If my calculations haven't gotten twisted, the answer should be approximately 8.63 meters.
Now, juggle these answers and make sure they don't go flying out of your hands!
To find the time it takes for the projectile to reach its maximum height, we can use the kinematic equation for vertical motion:
Δy = v0y * t - (1/2) * g * t^2
Where:
Δy = maximum height (unknown)
v0y = vertical component of initial velocity (unknown)
g = acceleration due to gravity (known as 9.8 m/s^2)
t = time (unknown)
Since the initial velocity is given as 9.2 m/s at an angle of 47°, we can find the vertical component of the initial velocity:
v0y = v0 * sin(θ)
v0y = 9.2 m/s * sin(47°)
Now we can substitute the given values into the equation and solve for t:
Δy = (9.2 m/s * sin(47°)) * t - (1/2) * (9.8 m/s^2) * t^2
Since the maximum height is reached when the projectile stops moving vertically (v = 0), we can set the vertical component of velocity equal to zero:
0 = (9.2 m/s * sin(47°)) - (9.8 m/s^2) * t
Solving this equation will give us the time it takes for the projectile to reach its maximum height.
To find the time it takes for the projectile to reach its maximum height, we can use the following equation of motion in the vertical direction:
Δy = v₀y * t + (1/2) * a * t²
where:
Δy is the maximum height,
v₀y is the vertical component of the initial velocity,
t is the time taken, and
a is the acceleration due to gravity.
We know that the initial velocity v₀ is 9.2 m/s and the angle of projection is 47°. We can find the vertical component of the initial velocity using trigonometry:
v₀y = v₀ * sin(θ)
where:
θ is the angle of projection.
Substituting the values:
v₀y = 9.2 m/s * sin(47°)
Now we can solve the equation for time t:
Δy = (9.2 m/s * sin(47°)) * t + (1/2) * (-9.8 m/s²) * t²
Since the projectile reaches its maximum height, the final vertical velocity is zero. So, we can find the time by setting the final velocity equal to zero:
0 = (9.2 m/s * sin(47°)) + (-9.8 m/s²) * t
Solving for t, we can rearrange the equation:
t = (9.2 m/s * sin(47°)) / 9.8 m/s²
Calculating the values using a calculator:
t ≈ 0.674 seconds (Answer to part a)
To find the time it takes for the projectile to hit the ground, we can use the equation of motion:
Δy = v₀y * t + (1/2) * a * t²
In this case, Δy would be the initial height of the cannon's barrel (6.6 m), and we need to solve for t. Since the final height is zero (projectile hits the ground), the equation becomes:
0 = (9.2 m/s * sin(47°)) * t + (1/2) * (-9.8 m/s²) * t²
Simplifying the equation and solving for t, we get:
4.9 * t² = (9.2 m/s * sin(47°)) * t
t² = (9.2 m/s * sin(47°)) / (4.9 m/s²)
Calculating the values:
t ≈ 0.175 seconds (Answer to part b)
To find the range Δx of the projectile, we can use the equation of motion in the horizontal direction:
Δx = v₀x * t
where:
Δx is the range,
v₀x is the horizontal component of the initial velocity, and
t is the time of flight.
The horizontal component of the initial velocity is given by:
v₀x = v₀ * cos(θ)
Substituting the values:
v₀x = 9.2 m/s * cos(47°)
Now we can calculate the range using the formula:
Δx = (9.2 m/s * cos(47°)) * t
Using the value of t calculated in part b:
Δx ≈ (9.2 m/s * cos(47°)) * 0.175 s
Δx ≈ 1.177 meters (Answer to part c)