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Holt Physics
Problem 3E
PROJECTILES LAUNCHED AT AN ANGLE
PROBLEM
The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The strait is only about 6.0 m wide. Suppose an athlete wanting to jump “over the sea” leaps at an angle of 35° with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance.
SOLUTION
1. DEFINE
Given:
∆x = 6.0 m q = 35° g = 9.81 m/s vi = ?
Unknown:
2. PLAN
Diagram:
v θ = 35°
∆x = 6.00 m
Choose the equation(s) or situation: The horizontal component of the athlete’s velocity, vx , is equal to the initial speed multiplied by the cosine of the angle, q, which is equal to the magnitude of the horizontal displacement, ∆x, divided by the time interval required for the complete jump.
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
∆x vx = vi cos q = ∆t At the midpoint of the jump, the vertical component of the athlete’s velocity, vy , which is the upward vertical component of the initial velocity, vi sin q, minus the downward component of velocity due to free-fall acceleration, equals zero. The time required for this to occur is half the time necessary for the total jump. ∆t vy = vi sin q − g = 0 2 g ∆t vi sin q = 2 Rearrange the equation(s) to isolate the unknown(s): Express ∆t in the second equation in terms of the displacement and velocity component in the first equation. g ∆x vi sin q = 2 vi cos q g ∆x vi 2 = 2 sin q cos q
24
Holt Physics Problem Workbook
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Lesson
NAME ______________________________________ DATE _______________ CLASS ____________________ g ∆x 2 sin q cos q
Print
vi =
3. CALCULATE
Substitute the values into the equation(s) and solve: Select the positive root for vi .
vi =
4. EVALUATE
NAME ______________________________________ DATE _______________ CLASS ____________________
Lesson
Holt Physics
Problem 3E
PROJECTILES LAUNCHED AT AN ANGLE
PROBLEM
The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The strait is only about 6.0 m wide. Suppose an athlete wanting to jump “over the sea” leaps at an angle of 35° with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance.
SOLUTION
1. DEFINE
Given:
∆x = 6.0 m q = 35° g = 9.81 m/s vi = ?
Unknown:
2. PLAN
Diagram:
v θ = 35°
∆x = 6.00 m
Choose the equation(s) or situation: The horizontal component of the athlete’s velocity, vx , is equal to the initial speed multiplied by the cosine of the angle, q, which is equal to the magnitude of the horizontal displacement, ∆x, divided by the time interval required for the complete jump.
Copyright © by Holt, Rinehart and Winston. Allrights reserved.
∆x vx = vi cos q = ∆t At the midpoint of the jump, the vertical component of the athlete’s velocity, vy , which is the upward vertical component of the initial velocity, vi sin q, minus the downward component of velocity due to free-fall acceleration, equals zero. The time required for this to occur is half the time necessary for the total jump. ∆t vy = vi sin q − g = 0 2 g ∆t vi sin q = 2 Rearrange the equation(s) to isolate the unknown(s): Express ∆t in the second equation in terms of the displacement and velocity component in the first equation. g ∆x vi sin q = 2 vi cos q g ∆x vi 2 = 2 sin q cos q
24
Holt Physics Problem Workbook
Menu
Lesson
NAME ______________________________________ DATE _______________ CLASS ____________________ g ∆x 2 sin q cos q
vi =
3. CALCULATE
Substitute the values into the equation(s) and solve: Select the positive root for vi .
vi =
4. EVALUATE