A projectile is launched off the roof of a 180-meter-tall building close to its edge, where the initial velocity's horizontal component is [tex]$20 m / s$[/tex] and its vertical component is [tex]$25 m / s$[/tex]. If the projectile lands on the ground below, how far from the building's side (in meters) does the projectile land? Round your answer to the nearest hundredth (0.01).
Hint: Find the time it spends in mid-air using the quadratic equation, then use that time to find the distance. If [tex]$A t^2+B t+C=0$[/tex], then [tex]$t=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}$[/tex]
It should give you two values for time; make sure to pick the right one!
Answer (2)
To find how far a projectile lands from a building, we first determine the time of flight with a quadratic equation, which gives us about 9.13 seconds. Using this time, we calculate the horizontal distance traveled as approximately 182.54 meters from the building. Thus, the projectile lands around 182.54 meters away.
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Determine the time the projectile spends in the air by solving the quadratic equation 4.9 t 2 − 25 t − 180 = 0 using the quadratic formula.
Choose the positive time value, t ≈ 9.1269 seconds.
Calculate the horizontal distance using the formula d i s t an ce = h or i zo n t a l v e l oc i t y × t im e = 20 × 9.1269 .
Round the final distance to the nearest hundredth: 182.54 meters.
Explanation
Problem Analysis We are given a projectile launched from a building with an initial height of 180 meters. The projectile has an initial horizontal velocity of 20 m/s and an initial vertical velocity of 25 m/s. We need to find the horizontal distance from the building where the projectile lands.
Finding the Time First, we need to find the time the projectile spends in the air. We can use the vertical motion to determine this. The equation for the vertical position is given by: h = v 0 t + 2 1 a t 2 where:
h is the final height (-180 m, since the projectile lands 180 m below the initial height)
v 0 is the initial vertical velocity (25 m/s)
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time
Setting up the Quadratic Equation Plugging in the values, we get: − 180 = 25 t − 4.9 t 2 Rearranging the equation, we have a quadratic equation: 4.9 t 2 − 25 t − 180 = 0
Solving for Time Now, we use the quadratic formula to solve for t :
t = 2 A − B ± B 2 − 4 A C where:
A = 4.9
B = -25
C = -180 Plugging in the values, we get: t = 2 ( 4.9 ) 25 ± ( − 25 ) 2 − 4 ( 4.9 ) ( − 180 ) t = 9.8 25 ± 625 + 3528 t = 9.8 25 ± 4153 t = 9.8 25 ± 64.44377 We have two possible values for t :
t 1 = 9.8 25 + 64.44377 = 9.8 89.44377 ≈ 9.1269 s t 2 = 9.8 25 − 64.44377 = 9.8 − 39.44377 ≈ − 4.0249 s Since time cannot be negative, we choose the positive value: t ≈ 9.1269 seconds.
Calculating the Horizontal Distance Next, we calculate the horizontal distance using the equation: d i s t an ce = h or i zo n t a l v e l oc i t y × t im e d i s t an ce = 20 × 9.1269 d i s t an ce = 182.538
Final Answer Rounding the distance to the nearest hundredth, we get: d i s t an ce ≈ 182.54 m e t ers
Examples
Understanding projectile motion is crucial in various real-world scenarios, such as in sports like baseball or golf, where calculating the trajectory of a ball helps athletes optimize their performance. Similarly, engineers use these principles to design systems ranging from launching satellites into orbit to predicting the landing points of emergency supplies dropped from an aircraft. By analyzing initial velocities, angles, and gravitational forces, one can accurately determine the range and flight path of projectiles, ensuring precision and efficiency in these applications.
