A ball is thrown vertically upward with a speed of +10.0 m/s.
(a) How high does it rise?
m
(b) How long does it take to reach its highest point?
s
(c) How long does the ball take to hit the ground after it reaches its highest point?
s
(d) What is its velocity when it returns to the level from which it started?
m/s
To find the answers to these questions, we can use the equations of motion for vertical motion. Here's how we can approach these questions step by step:
(a) To find how high the ball rises, we need to use the equation for displacement in vertical motion. The equation is given by:
Δy = v₀yt - (1/2)gt²
Where:
Δy is the vertical displacement (height),
v₀y is the initial vertical velocity,
t is the time, and
g is the acceleration due to gravity (approximately -9.8 m/s²).
In this case, the initial vertical velocity (v₀y) is +10.0 m/s and we assume the ball is thrown from the ground, so the initial vertical displacement (Δy) is 0.
Plugging in these values into the equation, we get:
0 = (10.0)(t) - (1/2)(9.8)(t²)
To solve for t, we need to rearrange the equation and solve for the time (t). The equation becomes a quadratic equation with the following form:
-4.9t² + 10.0t = 0
Solving this quadratic equation, we find that t = 0 s or t = 2.04 s. Since we are interested in the time it takes to reach the highest point, we discard the t = 0 s solution. Therefore, the ball takes 2.04 seconds to reach its highest point.
Now that we have the time it takes to reach the highest point, we can find the height (Δy) by substituting the value of t into the equation:
Δy = (10.0)(2.04) - (1/2)(9.8)(2.04)²
Calculating this, we find that Δy ≈ 20.8 meters. Therefore, the ball rises to a height of approximately 20.8 meters.
(b) The time it takes to reach the highest point is already calculated as 2.04 seconds.
(c) After the ball reaches its highest point, it will fall back down. The time it takes to hit the ground can be found using the equation:
Δy = v₀yt + (1/2)gt²
For the ball to hit the ground, the final vertical displacement (Δy) is -20.8 (since it falls back to the starting point). The initial vertical velocity (v₀y) is -10.0 m/s (since it's coming downward now), and we neglect the initial displacement (since it's falling from the same starting point).
Plugging in the values, we get:
-20.8 = (-10.0)(t) + (1/2)(-9.8)(t)²
Simplifying and rearranging the equation, we have:
4.9t² - 10.0t - 20.8 = 0
Solving this quadratic equation, we find that t ≈ 1.64 s or t ≈ 3.36 s. Since we are interested in the time it takes to hit the ground, we only consider the positive solution. Therefore, it takes approximately 3.36 seconds for the ball to hit the ground after reaching its highest point.
(d) Once the ball returns to the level from which it started, its velocity is the same as its original velocity, but in the opposite direction. Since the initial velocity was +10.0 m/s, the velocity when it returns will be -10.0 m/s.
Therefore, the velocity of the ball when it returns to the level from which it started is -10.0 m/s.