Question
A ball thrown with an initial velocity of 25.0 m/s [30.0°]. What is the maximum height?
a. 5.75 m
b. 6.85 m
c. 7.96 m
d. 8.35 m
a. 5.75 m
b. 6.85 m
c. 7.96 m
d. 8.35 m
Answers
GPT 3.5
The maximum height can be found using the following formula:
h = (vi^2 * sin^2(theta)) / (2 * g)
where:
h = maximum height
vi = initial velocity (25.0 m/s)
theta = launch angle (30.0°)
g = acceleration due to gravity (9.81 m/s^2)
Plugging in the values gives:
h = (25.0^2 * sin^2(30.0°)) / (2 * 9.81)
h = (625 * (0.5)^2) / 19.62
h = (625 * 0.25) / 19.62
h = 156.25 / 19.62
h = 7.96 m
Therefore, the maximum height is approximately 7.96 m, so the correct answer is option c.
h = (vi^2 * sin^2(theta)) / (2 * g)
where:
h = maximum height
vi = initial velocity (25.0 m/s)
theta = launch angle (30.0°)
g = acceleration due to gravity (9.81 m/s^2)
Plugging in the values gives:
h = (25.0^2 * sin^2(30.0°)) / (2 * 9.81)
h = (625 * (0.5)^2) / 19.62
h = (625 * 0.25) / 19.62
h = 156.25 / 19.62
h = 7.96 m
Therefore, the maximum height is approximately 7.96 m, so the correct answer is option c.