A baseball is thrown at an angle of 23◦ relative to the ground at a speed of 22.3 m/s. The ball is caught 36.4649 m from the thrower. The acceleration due to gravity is 9.81 m/s2 . How long is it in the air?
How high is the tallest spot in the ball’s path?
Horizontal velocity component = 22.3 cos23 = 20.53 m/s
Initial horizontal velocity component
Vyo = 22.3 sin23 = 8.71 m/s
Time ball is in the air = 36.4649/20.53 = 1.78 s.
I did not have to use the horizontal velocity component to get that result; one can use Voy to get the time in the air as follows:
T = 2*Vyo/g = 1.78 s
Both methods agree, as they should.
a person standing at the edge of a seaside cliff kick is 51 m above the water's surface, as shown. the acceleration of gravity is 9.81 m/s^2
a) how long does it take for the stone to fall to the water?
b) with what speed does the stone strike the water.
To find out how long the baseball is in the air, we can use the fact that the vertical motion of the ball follows a projectile motion trajectory.
First, let's break down the initial velocity of the ball into its horizontal and vertical components. We can use trigonometry to find these components:
Vertical component: V_y = V * sin(theta)
V_y = 22.3 m/s * sin(23°)
V_y ≈ 8.55 m/s
Horizontal component: V_x = V * cos(theta)
V_x = 22.3 m/s * cos(23°)
V_x ≈ 20.45 m/s
Since the ball lands at the same height it was thrown (assuming the ground is level), the total time in the air can be found by dividing the horizontal distance traveled (36.4649 m) by the horizontal velocity (V_x):
Time in air = Distance / Horizontal velocity
Time in air = 36.4649 m / 20.45 m/s
Time in air ≈ 1.78 seconds
To find the maximum height of the ball's path, we can determine the vertical distance traveled using the same time of flight (1.78 seconds) and the vertical velocity component (V_y):
Vertical distance = V_y * Time in air - (1/2) * g * (Time in air)^2
Vertical distance = 8.55 m/s * 1.78 s - (1/2) * 9.81 m/s^2 * (1.78 s)^2
Vertical distance ≈ 7.6 meters
Therefore, the highest point in the ball's path is approximately 7.6 meters above the initial throwing height.
To find the time it takes for the baseball to be in the air, we can use the equation of motion for an object in projectile motion:
𝑑 = 𝑢𝑡 + (1/2)𝑎𝑡^2
Where:
𝑑 is the horizontal distance traveled by the baseball
𝑢 is the initial horizontal velocity of the baseball (the horizontal component of the velocity)
𝑡 is the time taken
𝑎 is the horizontal acceleration (which is 0 in this case)
We have 𝑢 and 𝑑 from the given information, so we can rearrange the equation to solve for 𝑡:
𝑑 = 𝑢𝑡
𝑡 = 𝑑 / 𝑢
Substituting the values:
𝑡 = 36.4649 m / 22.3 m/s
𝑡 ≈ 1.6365 s
Therefore, the baseball is in the air for approximately 1.6365 seconds.
To find the highest point in the ball's path, we need to consider the vertical motion. The vertical component of the velocity at any given time can be calculated using the equation:
𝑉𝑦 = 𝑢𝑦 + 𝑎𝑦𝑡
Where:
𝑉𝑦 is the vertical component of the velocity
𝑢𝑦 is the initial vertical velocity of the baseball
𝑎𝑦 is the vertical acceleration (due to gravity, which is -9.81 m/s^2)
𝑡 is the time taken
At the highest point, the vertical component of the velocity becomes 0. Therefore, we can use this equation to find the time it takes for the ball to reach the highest point:
0 = 𝑢𝑦 + 𝑎𝑦𝑡
Rearranging the equation:
𝑡 = -𝑢𝑦 / 𝑎𝑦
To find 𝑢𝑦, we can use the given information that the ball is thrown at an angle of 23° relative to the ground and a speed of 22.3 m/s. The initial vertical velocity can be calculated using trigonometry:
𝑢𝑦 = 𝑢 * sin(𝑡ℎ𝑟𝑜𝑤𝑛 𝑎𝑛𝑔𝑙𝑒)
𝑢𝑦 = 22.3 m/s * sin(23°)
𝑢𝑦 ≈ 8.29 m/s
Substituting the values:
𝑡 = -8.29 m/s / (-9.81 m/s^2)
𝑡 ≈ 0.845 s
Therefore, it takes approximately 0.845 seconds for the ball to reach the highest point.
Now, we can use the time to calculate the maximum height using the equation of motion:
𝑑 = 𝑢𝑦𝑡 + (1/2)𝑎𝑦𝑡^2
Rearranging the equation:
𝑑 = 𝑢𝑦 * 𝑡 + (1/2)𝑎𝑦𝑡^2
Substituting the values:
𝑑 = 8.29 m/s * 0.845 s + (1/2) * (-9.81 m/s^2) * (0.845 s)^2
𝑑 ≈ 3.62 m
Therefore, the highest spot in the ball's path is approximately 3.62 meters.