A ball is thrown straight upward and returns to the thrower's hand after 2.00 s in the air. A second ball thrown at an angle of 40.0° with the horizontal reaches the same maximum height as the first ball.
(a) At what speed was the first ball thrown?
m/s
(b) At what speed was the second ball thrown?
m/s
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
s
See previous post: Wed, 2-3-16, 11:49 AM.
To answer these questions, we'll use the equations of motion for projectile motion. We'll assume that the upward direction is positive and that there is no air resistance.
First, let's determine the time it takes for the first ball to reach its maximum height.
Using the formula for the time of flight given an initial vertical velocity (u) and an acceleration due to gravity (g):
t = u/g
Since the ball is thrown straight upward, the initial vertical velocity (u) is equal to the final vertical velocity (v) when the ball returns to the thrower's hand, and the time of flight is given as 2.00 s.
Thus, 2.00 s = u/g
Now, let's find the value of g. The acceleration due to gravity is approximately 9.8 m/s^2.
Substituting the value of g into the equation:
2.00 s = u/(9.8 m/s^2)
Solving for u:
u = 2.00 s * 9.8 m/s^2
(a) The speed (u) at which the first ball was thrown is approximately 19.6 m/s.
Next, let's determine the speed at which the second ball was thrown, given that it reaches the same maximum height as the first ball.
When considering projectile motion at an angle, we can break the initial velocity (u) into its horizontal (u_x) and vertical (u_y) components.
The vertical component can be determined using the formula:
u_y = u * sin(θ)
where θ is the angle with the horizontal (40 degrees in this case).
Now, let's find the value of u_y.
u_y = u * sin(40°)
Since the maximum height reached by the second ball is the same as the first ball, the vertical component of the second ball's initial velocity (u_y) is equal to the vertical final velocity when it reaches its maximum height.
Using the equation of motion:
v = u + at
The final vertical velocity (v) is 0 m/s when the ball reaches its maximum height because it momentarily stops moving vertically before falling back down.
So, 0 m/s = u_y - g * t_max_height
Solving for u_y:
u_y = g * t_max_height
Substituting in the values of g (9.8 m/s^2) and t_max_height (which we found to be 2.00 s in the previous part):
u_y = 9.8 m/s^2 * 2.00 s
Now, let's find the horizontal component of the initial velocity (u_x) using the formula:
u_x = u * cos(θ)
Substituting the values of u (which we found to be 19.6 m/s) and θ (40 degrees):
u_x = 19.6 m/s * cos(40°)
Finally, we can determine the speed (u) at which the second ball was thrown using the Pythagorean theorem:
u = sqrt(u_x^2 + u_y^2)
(b) The speed (u) at which the second ball was thrown is approximately 19.0 m/s.
Lastly, let's determine the time it takes for the ball to reach a point 10.0 m below the level of launching.
We'll use the formula for the displacement in the vertical direction:
h = u_y * t + 0.5 * g * t^2
Since we want to find the time it takes for the ball to reach a point 10.0 m below the level of launching, we can set h equal to -10.0 m (negative because it is below the level of launching).
-10.0 m = u_y * t - 0.5 * g * t^2
Rearranging the equation:
0.5 * g * t^2 - u_y * t - 10.0 m = 0
This is a quadratic equation that can be solved using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where a = 0.5 * g, b = -u_y, and c = -10.0 m.
Substituting in the values of g (9.8 m/s^2), u_y (which we found previously), and c (-10.0 m):
t = (- (-u_y) ± sqrt((-u_y)^2 - 4 * 0.5 * 9.8 * (-10.0))) / (2 * 0.5 * 9.8)
(c) The time it takes for the ball to reach a point 10.0 m below the level of launching is approximately 2.43 s.
To solve these problems, we can use the equations of motion for projectile motion. In projectile motion, the vertical motion follows the equation:
y = u*t - (1/2)*g*t^2,
where y is the vertical displacement, u is the initial vertical velocity, t is the time, and g is the acceleration due to gravity.
In this case, we know that the first ball is thrown straight upward and returns to the thrower's hand after 2.00 s. This means the total time of flight is 2.00 s.
(a) To find the initial speed (u) of the first ball, we need to determine the maximum height reached and use the equation for vertical displacement:
y = u*t - (1/2)*g*t^2.
Since the ball reaches the same maximum height, the vertical displacement at the maximum height for both balls is the same.
Let's call the maximum height H. So, for the first ball:
H = u*t - (1/2)*g*t^2.
Substituting the given values, we get:
H = u*2.00 - (1/2)*9.81*(2.00)^2.
Now, let's determine the maximum height H.
H = (u*2.00) - (1/2)*(9.81)*(2.00)^2
We know that the maximum height for both balls is the same, so we can use this value for the second ball as well.
(b) For the second ball, we can use the same equation for vertical displacement and solve for the initial speed (u):
H = u*t - (1/2)*g*t^2.
Substituting the given angle and its components:
H = (u*sin(40))*t - (1/2)*g*t^2,
where u*sin(40) is the initial vertical velocity component.
(c) For part (c), we need to determine the time it takes for the ball to reach a point 10.0 m below the launching level. Again, we can use the equation for vertical displacement:
y = u*t - (1/2)*g*t^2.
Since the vertical displacement is given as negative (below the launching level), we can set:
-10.0 = u*t - (1/2)*g*t^2,
and solve for t.
Once we have the values for H in parts (a) and (b), and t for part (c), we can calculate the speeds of the balls using the horizontal motions. However, we need the values for H and t to proceed further with the calculations.