A projectile of mass 0.539 kg is shot from a
cannon. The end of the cannon’s barrel is
at height 6.4 m, as shown in the figure. The
initial velocity of the projectile is 9 m/s .
How long does it take the projectile to hit the
ground?
Answer in units of s
Well, let me clown around with the calculations for a bit. In order to find the time it takes for the projectile to hit the ground, we can use the classic equation of motion:
h = Vit + (1/2)at^2
Given that the initial velocity (Vi) is 9 m/s and the height (h) is 6.4 m, we can plug in these values:
6.4 = 9t + (1/2)(9.8)t^2
Now comes the funny part: solving this quadratic equation. Time for some clown magic!
*tosses juggling balls, twirls a horn*
After using my magical clown calculator, I found that the equation simplifies to:
4.9t^2 + 9t - 6.4 = 0
And... drumroll, please!
Using the quadratic formula, the two solutions I got are: t = -1.14 s and t = 0.67 s.
Now, I know being negative is not cool, so let's remove the negative one from the equation, and the final answer is:
t = 0.67 s
So, the projectile will take approximately 0.67 seconds to hit the ground! Easy peasy, lemon squeezy!
To find the time it takes for the projectile to hit the ground, we can use the equations of motion.
First, let's consider the vertical motion of the projectile. The initial height is given as 6.4 m, and we need to find the time it takes for the projectile to reach the ground. We can use the equation:
h = ut + (1/2)gt^2
Where:
h = height (6.4 m)
u = initial velocity (9 m/s)
g = acceleration due to gravity (-9.8 m/s^2, considering downward direction)
t = time
Substituting the given values into the equation:
6.4 = (9)(t) + (1/2)(-9.8)(t^2)
Rearranging the equation:
4.9t^2 + 9t - 6.4 = 0
Now we can solve this quadratic equation to find the value of t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
Where:
a = 4.9
b = 9
c = -6.4
Substituting the values into the quadratic formula:
t = (-9 ± √(9^2 - 4(4.9)(-6.4)))/(2(4.9))
Calculating the discriminant:
√(9^2 - 4(4.9)(-6.4)) = √(81 + 126.56) = √207.56 = 14.4
Substituting this value into the equation:
t = (-9 ± 14.4)/(2(4.9))
Now we have two possible values for t:
t1 = (-9 + 14.4)/(2(4.9)) = 0.918 s
t2 = (-9 - 14.4)/(2(4.9)) = -2.5 s
Since time cannot be negative in this context, we discard the negative value. Therefore, the time it takes for the projectile to hit the ground is approximately 0.918 seconds.
To calculate the time it takes for the projectile to hit the ground, we need to use the principles of projectile motion and gravitational acceleration.
First, let's define the known values:
Mass of the projectile (m) = 0.539 kg
Initial velocity (v₀) = 9 m/s
Height of the cannon's barrel (h) = 6.4 m
Assuming no air resistance, the vertical motion of the projectile is purely influenced by the force of gravity.
We can use the kinematic equation that relates the vertical position, initial velocity, time, and acceleration:
h = v₀t + (1/2)gt²
Where:
h is the height (6.4 m)
v₀ is the initial velocity (9 m/s)
g is the acceleration due to gravity (9.8 m/s²)
t is the time it takes for the projectile to hit the ground (what we want to find)
Since the projectile is hitting the ground, the final height (h) will be zero. Therefore, we can rewrite the equation as:
0 = 9t + (1/2)(9.8)t²
Now, let's solve the quadratic equation for t. Multiplying through by 2 to eliminate the fraction, we have:
0 = 18t + 9.8t²
Rearranging the equation to standard quadratic form:
9.8t² + 18t = 0
Factoring out t:
t(9.8t + 18) = 0
Setting each factor equal to zero:
t = 0 or 9.8t + 18 = 0
Since time cannot be negative, we discard the t = 0 solution.
Solving for t in the second equation:
9.8t = -18
t = -18 / 9.8
t ≈ -1.84 s
Since time cannot be negative in this context, we disregard this negative value.
Therefore, the time it takes for the projectile to hit the ground is approximately 1.84 seconds.