Question

2. A locust jumps at an angle of 55.0° and lands 0.800 m from where it jumped. What is the maximum height of the locust during its jump? (Keep in mind that vi is not known. You will need to come up with 2 equations with 2 unknowns. One idea is to solve for the total time first, then vi, and finally ymax). Projectile motion : ?x = vix ?t yf = yi + viy ?t + 1/2g (?t)² vfy = viy + g ?t vfy² = viy² + 2g(yf - yi) (g = -9.8 m/s²)

          2. A locust jumps at an angle of 55.0° and lands 0.800 m from where it jumped. What is the maximum height of the locust during its jump? (Keep in mind that vi is not known. You will need to come up with 2 equations with 2 unknowns. One idea is to solve for the total time first, then vi, and finally ymax).

Projectile motion : ?x = vix ?t
yf = yi + viy ?t + 1/2g (?t)²
vfy = viy + g ?t
vfy² = viy² + 2g(yf - yi) (g = -9.8 m/s²)

2. A locust jumps at an angle of 55.0° and lands 0.800 m from where it jumped. What is the maximum height of the locust during its jump? (Keep in mind that vi is not known. You will need to come up with 2 equations with 2 unknowns. One idea is to solve for the total time first, then vi, and finally ymax).

Projectile motion : ?x = vix ?t
yf = yi + viy ?t + 1/2g (?t)²
vfy = viy + g ?t
vfy² = viy² + 2g(yf - yi) (g = -9.8 m/s²)

Added by Melinda R.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Penny Riley

12/05/2022

Step 1

We are given the angle of projection (55.0°), the horizontal distance covered during the jump (0.800 m), and the acceleration due to gravity (g = 9.8 m/s²). We need to find the maximum height (Ymax) of the locust during its jump. Show more…

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A locust jumps at an angle of 55.0° and lands 0.800 m from where it jumped. What is the maximum height of the locust during its jump? (Keep in mind that vi is not known. You will need to come up with 2 equations with 2 unknowns. One idea is to solve for the total time first, then vi, and finally ymax.) Projectile motion : Δx = vix Δt yf = yi + viy Δt + 1/2g (Δt)² vfy = viy + g Δt vfy² = viy² + 2g(yf - yi) (g = -9.8 m/s²)