A ball is thrown vertically upward with a speed of +13.0 m/s.
(a) How high does it rise? (b) How long does it take to reach its highest point? (c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?
a. h = (Vf^2 - Vo^2) / 2g,
h = (0 - (13)^2) / -19.6 = 8.62m.
b. t(up) = (Vf - Vo) / g,
t(up) = (0 - 13) / -9.8 = 1.33s.
c. d = Vo*t + 4.9t^2 = 8.62,
13t + 4.9t^2 = 8.62,
4.9t^2 + 13t - 8.62 = 0Use Quadratic Formula to find t:
t = 0.55s.
d. Vf^2 = Vo^2 + 2gd,
Vf^2 = 0 + 2*9.8*8.62 = 168.952,
Vf = 13m/s. = Same as initial velocity
going up.
(a) The ball rises so high that it even reached out to the heavens inviting Zeus for a game of catch. Just kidding! To find the height it rises, we can use the equation for vertical displacement: h = (v^2 - u^2) / 2g, where h is the height, v is the final velocity, u is the initial velocity, and g is the acceleration due to gravity. Plugging in the values, we get h = (0 - 13^2) / (2 * (-9.8)) = 8.42 meters (rounded to two decimal places).
(b) The time it takes to reach the highest point can be found using the equation: t = (v - u) / g, where t is the time, v is the final velocity, u is the initial velocity, and g is the acceleration due to gravity. Plugging in the values, we get t = (0 - 13) / (-9.8) = 1.33 seconds (rounded to two decimal places).
(c) The time it takes for the ball to hit the ground after reaching its highest point is equal to twice the time it took to reach the highest point. So, the time is 2 * 1.33 = 2.66 seconds (rounded to two decimal places).
(d) When the ball returns to the level from which it started, its velocity is equal in magnitude but opposite in direction to its initial velocity. So, the velocity is -13.0 m/s.
Hope that helps!
To answer these questions, we can use the equations of motion for a vertically thrown object under gravity. Let's assume the acceleration due to gravity is -9.8 m/s² (taking downward as negative).
(a) To find the height the ball rises, we can use the equation:
v² = u² + 2as
where:
v = final velocity (0 m/s when it reaches the highest point)
u = initial velocity (+13.0 m/s)
a = acceleration (-9.8 m/s²)
s = displacement or height
Rearranging the equation, we have:
s = (v² - u²) / (2a)
Plugging in the values, we get:
s = (0² - (+13.0)²) / (2*(-9.8))
= (-169) / (-19.6)
= 8.63 meters
Therefore, the ball rises to a height of approximately 8.63 meters.
(b) To find the time it takes to reach its highest point, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s when it reaches the highest point)
u = initial velocity (+13.0 m/s)
a = acceleration (-9.8 m/s²)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (0 - (+13.0)) / (-9.8)
= -13.0 / (-9.8)
= 1.33 seconds
Therefore, it takes approximately 1.33 seconds to reach the highest point.
(c) When the ball reaches its highest point, its velocity is momentarily zero before gravity starts pulling it downward. The time it takes to fall back to the ground is the same as the time it takes to reach the highest point, which is 1.33 seconds.
(d) To find the velocity when the ball returns to the level from which it started, we can use the equation:
v = u + at
where:
v = final velocity
u = initial velocity (+13.0 m/s)
a = acceleration (-9.8 m/s²)
t = time
Plugging in the values, we get:
v = (+13.0) + (-9.8) * (1.33)
= +13.0 - 13.03
= -0.03 m/s
Therefore, the ball's velocity when it returns to the level from which it started is approximately -0.03 m/s (negative, indicating downward direction).
To solve this problem, we can use the equations of motion for projectile motion. Let's break the problem down and find the answers step by step.
(a) How high does it rise?
The ball is thrown upward, so the initial velocity is +13.0 m/s. The acceleration due to gravity is -9.8 m/s² (taking downward as negative). We can use the equation:
v² = u² + 2as
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (+13.0 m/s)
a = acceleration (-9.8 m/s²)
s = displacement
Rearranging the equation, we have:
s = (v² - u²) / (2a)
Substituting the given values, we get:
s = (0 - (13.0)²) / (2 * -9.8)
Calculating this gives us:
s = -21.023 m
Since the displacement is negative, it means the ball rises 21.023 meters.
(b) How long does it take to reach its highest point?
We can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (+13.0 m/s)
a = acceleration (-9.8 m/s²)
t = time
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0 - 13.0) / (-9.8)
Calculating this gives us:
t = 1.327 s
Therefore, it takes approximately 1.327 seconds for the ball to reach its highest point.
(c) How long does the ball take to hit the ground after it reaches its highest point?
The time it takes for an object to fall from a certain height can be determined using the equation:
s = ut + 1/2at²
In this case, the ball starts falling from the highest point, and its initial velocity is 0 m/s. So we have:
s = 0*t + 1/2*(-9.8)*t²
Simplifying this equation, we get:
4.9t² = s
where s is the height the ball rises from part (a).
Substituting the value of s = -21.023 m (from part (a)), we get:
4.9t² = -21.023
t² = -4.301
Since time cannot be negative, we disregard the negative value.
Therefore, the ball takes approximately 4.301 seconds to hit the ground after it reaches its highest point.
(d) What is its velocity when it returns to the level from which it started?
To find the velocity when the ball returns to its starting level, we use the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the ball returns to its starting level, the displacement is 0. Therefore, the final velocity of the ball when it returns is equal to the negative of its initial velocity.
Substituting the given values, we have:
v = (-13.0) + (-9.8)(4.301)
Calculating this gives us:
v = -55.701 m/s
Therefore, the velocity of the ball when it returns to the level from which it started is approximately -55.701 m/s.