At t = 0, an arrow is fired vertically upwards with a speed of 100 m/s. A second arrow is fired vertically upwards with the same speed at t = 5 s. ThenSelect answersAThe two arrows will be at the same height above the ground at t = 5 s. ThenBThe two arrows will reach back their starting points at t = 20 s and at t = 25 s.CThe ratio of the speeds of the first and second arrows at t = 20 s will be 2 : 1DThe maximum height attained by either arrow will be 1000 m

An arrow is shot straight up in the air at an initial speed of 15.0 m/s. After how much time is the arrow moving downward at a speed of 8.00 m/s? 2 points3.22 s0.714 s2.35 s1.87 s1.24 s

An arrow is shot into the air and its height in feet after t𝑡 seconds is given by the function f(t)=−16t2+128t𝑓(𝑡)=−16𝑡2+128𝑡. Estimate the instantaneous velocity at t=1𝑡=1 second using difference quotients with h=0.1ℎ=0.1, 0.010.01, and 0.0010.001. If necessary, round the difference quotients to no less than six decimal places and round your final answer to the nearest integer.

A projectile is shot vertically upward with some initial velocity.It reaches a maximum height of 100 m.If, on a second shot, the initial velocity is doubled, what will the new maximum height be?Neglect friction and express your answer in meters.

Let’s revisit our archery problem from earlier with a little more information.Andrew is an avid archer. He launches an arrow that takes a parabolic path. The equation of the height of the arrow with respect to time is y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥, where y𝑦 is the height of the arrow in meters above Andrew’s bow and x𝑥 is the time in seconds since Andrew shot the arrow. Find how long it takes the arrow to come back to a height even with his bow height.Let’s graph the equation by making a table of values. Remember it is helpful to start with finding the vertex so that we can know what values to use for x𝑥.y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥Find the x𝑥-value of the vertex by using the formula −b2a−𝑏2𝑎.−482(−4.9)≈4.9−482(−4.9)≈4.9Note: We used the ≈≈ symbol instead of the == sign because we are having to round this value. Remember we aren’t going to get as many “pretty” numbers with real world problems.)Now we’ll substitute that into our equation to find the y𝑦-value of our vertex.y=−4.9(𝑦=−4.9( )2+48()2+48( )=117.6)=117.6Our vertex is approximately (( ,, )).We’ll create our table of values centered around our vertex (but will round to 55 for that spot in our table).x𝑥y=−4.9x2+48x𝑦=−4.9𝑥2+48𝑥00y=−4.9(0)2+48(0)=0𝑦=−4.9(0)2+48(0)=011y=−4.9(1)2+48(1)=43.1𝑦=−4.9(1)2+48(1)=43.122y=−4.9(2)2+48(2)=76.4𝑦=−4.9(2)2+48(2)=76.433y=−4.9(3)2+48(3)=99.9𝑦=−4.9(3)2+48(3)=99.944y=−4.9(4)2+48(4)=113.6𝑦=−4.9(4)2+48(4)=113.655y=−4.9(5)2+48(5)=117.6𝑦=−4.9(5)2+48(5)=117.666y=−4.9(6)2+48(6)=111.6𝑦=−4.9(6)2+48(6)=111.677y=−4.9(7)2+48(7)=95.9𝑦=−4.9(7)2+48(7)=95.988y=−4.9(8)2+48(8)=70.4𝑦=−4.9(8)2+48(8)=70.499y=−4.9(9)2+48(9)=35.1𝑦=−4.9(9)2+48(9)=35.11010y=−4.9(10)2+48(10)=−10𝑦=−4.9(10)2+48(10)=−10Here’s the graph of the function.The roots of the function are approximately x=0𝑥=0 sec and x=9.8𝑥=9.8 sec. The first root tells us that the height of the arrow was 00 meters above his bow after seconds (right before Andrew first shot it) which makes sense. The second root says that it takes approximately seconds for the arrow to return to the height of the bow.Remember that we already calculated our vertex, so we don’t need to inspect our graph to find it, but we can use it to confirm it. We can interpret our vertex to mean that at approximately seconds, the arrow reached a maximum/minimum height of meters.CheckQuestion 7

An archer puts a 0.30kg arrow to the bowstring. An average force of 201N is exerted to draw the string back 0.8m. a. Assuming no frictional loss, with what speed does the arrow leave the bow? b. If the arrow is shot vertically into the air, how high will it rise?