A projectile is launched off the roof of a 200-meter-tall building close to its edge, where the initial velocity's horizontal component is [tex]$10 m / s$[/tex] and its vertical component is [tex]$15 m / s$[/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]$t=\frac{-B+\sqrt{B^2-4 A C}}{2 A}$[/tex]
It should give you two values for time; make sure to pick the right one!
Answer (1)
Find the time the projectile spends in the air using the vertical motion and the equation Δ y = v 0 y t + 2 1 a t 2 .
Solve the quadratic equation 4.9 t 2 − 15 t − 200 = 0 using the quadratic formula to find the time t ≈ 8.100 seconds.
Calculate the horizontal distance using the horizontal velocity and time: x = v 0 x t = 10 × 8.100 = 81.00 meters.
The projectile lands approximately 81.00 meters from the base of the building.
Explanation
Problem Analysis We are given a projectile launched from a 200-meter-tall building with initial horizontal velocity 10 m / s and initial vertical velocity 15 m / s . We need to find the horizontal distance from the building where the projectile lands.
Finding the Time in the Air First, we need to find the time the projectile spends in the air. We can use the vertical motion to determine this. The vertical displacement is -200 meters (since the projectile lands 200 meters below the initial height). We use the equation of motion: Δ y = v 0 y t + 2 1 a t 2 , where Δ y = − 200 , v 0 y = 15 , and a = − 9.8 m / s 2 (acceleration due to gravity).
Setting up the Quadratic Equation Plugging in the values, we get − 200 = 15 t + 2 1 ( − 9.8 ) t 2 , which simplifies to − 200 = 15 t − 4.9 t 2 . Rearranging the equation, we have 4.9 t 2 − 15 t − 200 = 0 . This is a quadratic equation in the form of A t 2 + Bt + C = 0 , where A = 4.9 , B = − 15 , and C = − 200 .
Applying the Quadratic Formula We can use the quadratic formula to solve for t : t = 2 A − B ± B 2 − 4 A C . Plugging in the values, we get t = 2 ( 4.9 ) − ( − 15 ) ± ( − 15 ) 2 − 4 ( 4.9 ) ( − 200 ) = 9.8 15 ± 225 + 3920 = 9.8 15 ± 4145 .
Calculating Time Values Calculating the two possible values for t , we have t 1 = 9.8 15 + 4145 ≈ 9.8 15 + 64.3817 ≈ 8.100 and t 2 = 9.8 15 − 4145 ≈ 9.8 15 − 64.3817 ≈ − 5.039 . Since time cannot be negative, we choose the positive root, t ≈ 8.100 seconds.
Calculating Horizontal Distance Now, we can calculate the horizontal distance using the horizontal velocity and the time. The horizontal distance x is given by x = v 0 x t , where v 0 x = 10 m / s . Therefore, x = 10 × 8.100 = 81.00 meters.
Final Answer Rounding the horizontal distance to the nearest hundredth, we get 81.00 meters.
Examples
Understanding projectile motion is crucial in various real-world scenarios, such as in sports like baseball or basketball, where calculating the trajectory of a ball helps players make accurate throws or shots. Similarly, in engineering, understanding projectile motion is essential for designing systems that involve launching objects, like in delivery systems or even in understanding the path of water in irrigation systems. By understanding the principles of projectile motion, we can predict and control the movement of objects in a variety of applications.
