13. Please answer this ASAP. It's pyshics. **Problem 13:**Sally Science hits a golf shot that travels at 31.88 m/s. The shot strikes a tree and bounces back to the tee at 7.97 m/s (exactly one-fourth the initial speed). The round trip from tee to tree to tee again took exactly 3.82 seconds. How far away is the tree from the tee?---### Explanation:In this problem, we need to calculate the distance the golf ball travels to the tree and back, given its speeds and total travel time. Here is a breakdown of the approach:1. **Initial Speed to the Tree**: 31.88 m/s2. **Speed Back to the Tee**: 7.97 m/s (one-fourth of the initial speed)3. **Total Round Trip Time**: 3.82 secondsLet's denote the distance to the tree as \( d \).- **Time to the Tree**: \( \frac{d}{31.88} \)- **Time Back to the Tee**: \( \frac{d}{7.97} \)The sum of these times is the total round trip time:\[\frac{d}{31.88} + \frac{d}{7.97} = 3.82\]Solving this equation will give us the distance \( d \).

Name: 13. Please answer this ASAP. It's pyshics. **Problem 13:**Sally Scienc
Uploaded: 2024-03-05T11:56:34.000Z
Channel: Bartleby
Description: 13. Please answer this ASAP. It's pyshics. **Problem 13:**Sally Science hits a golf shot that travels at 31.88 m/s. The shot strikes a tree and bounces back to the tee at 7.97 m/s (exactly one-fourth the initial speed). The round trip from tee to tre

Given data: The initial speed of the golf shot is u=31.88 m/s The final speed of the golf shot is v=7.97 m/s Let the distance between the tee to tree is d; The distance in the both the cases is same, only the time varies; Applying the time, distance and speed relation; distanced=speedu×timetSubstitute the initial speed; d=31.88×t1 Substitute the final speed; d=7.97×t2 Since the distance is same; 31.88×t1=7.97×t2 The total time is t1+t2=3.82 s So; 31.88×t1=7.97×t231.88×3.82-t2=7.97×t2121.78-31.88t2=7.97t2t2=3 s The distance d is; d=7.97×t2=7.97×3=23.9 m Thus, the distance of the tee from the tree is 23.9 m.