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Solution for (a)

By “height” we mean the altitude or vertical position y size 12{y} {} above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0 size 12{ v rSub { size 8{y} } =0} {} . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y size 12{y} {} :

v y 2 = v 0 y 2 2 g ( y y 0 ) . size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) "."} {}
The x y graph shows the trajectory of fireworks shell. The initial velocity of the shell v zero is at angle theta zero equal to seventy five degrees with the horizontal x axis. The fuse is set to explode the shell at the highest point of the trajectory which is at a height h equal to two hundred thirty three meters and at a horizontal distance x equal to one hundred twenty five meters from the origin.
The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 size 12{y rSub { size 8{0} } } {} and v y size 12{v rSub { size 8{y} } } {} are both zero, the equation simplifies to

0 = v 0 y 2 2 gy. size 12{0=v rSub { size 8{0y} } rSup { size 8{2} } - 2 ital "gy."} {}

Solving for y size 12{y} {} gives

y = v 0 y 2 2 g . size 12{y= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } "." } {}

Now we must find v 0 y size 12{v rSub { size 8{0y} } } {} , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ size 12{v rSub { size 8{0y rSup} =v rSub {0 rSup size 12{"sin"θ}} {} , where v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º size 12{θ rSub { size 8{0} } } {} is the initial angle. Thus,

v 0 y = v 0 sin θ 0 = ( 70.0 m/s ) ( sin 75º ) = 67.6 m/s. size 12{v rSub { size 8{0y} } =v rSub { size 8{0} } "sin"θ rSub { size 8{0} } = \( "70" "." 0" m/s" \) \( "sin""75" { size 12{ circ } } \) ="67" "." 6" m/s."} {}

and y size 12{y} {} is

y = ( 67 .6 m/s ) 2 2 ( 9 . 80 m /s 2 ) , size 12{y= { { \( "67" "." 6" m/s" \) rSup { size 8{2} } } over {2 \( 9 "." "80"" m/s" rSup { size 8{2} } \) } } } {}

so that

y = 233 m. size 12{y="233"" m."} {}

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {} . Because y 0 size 12{y rSub { size 8{0} } } {} is zero, this equation reduces to simply

y = 1 2 ( v 0 y + v y ) t . size 12{y= { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}

Note that the final vertical velocity, v y size 12{v rSub { size 8{y} } } {} , at the highest point is zero. Thus,

t = 2 y ( v 0y + v y ) = 2 ( 233 m ) ( 67.6 m/s ) = 6.90 s . alignl { stack { size 12{t= { {2y} over { \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) } } = { {2 times "233"" m"} over { \( "67" "." 6" m/s" \) } } } {} #=6 "." "90"" s" {} } } {}

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using y = y 0 + v 0 y t 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {} , and solving the quadratic equation for t size 12{t} {} .)

Solution for (c)

Because air resistance is negligible, a x = 0 size 12{a rSub { size 8{x} } =0} {} and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {} , where x 0 size 12{x rSub { size 8{0} } } {} is equal to zero:

x = v x t , size 12{x=v rSub { size 8{x} } t ","} {}

where v x size 12{v rSub { size 8{x} } } {} is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 8{0} } "." } {} Now,

v x = v 0 cos θ 0 = ( 70 . 0 m/s ) ( cos 75.0º ) = 18 . 1 m/s. size 12{v rSub { size 8{x} } =v rSub { size 8{0} } "cos"θ rSub { size 12{0} } = \( "70" "." 0" m/s" \) \( "cos""75.0º" \) ="18" "." 1" m/s."} {}

The time t size 12{t} {} for both motions is the same, and so x size 12{t} {} is

x = ( 18 . 1 m/s ) ( 6 . 90 s ) = 125 m. size 12{x= \( "18" "." 1" m/s" \) \( 6 "." "90"" s" \) ="125"" m."} {}

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y size 12{y} {} is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h size 12{y=h} {} ; then,

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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Source:  OpenStax, Cc test coll. OpenStax CNX. Dec 15, 2015 Download for free at http://legacy.cnx.org/content/col11717/1.4
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