A baseball is thrown at an angle of 25degrees relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air?
How high above the thrower did the ball travel?
This is essentially a vector problem where you are given the resultant vector and have to find the x- and y- components of the vector. When you throw a ball into the air at an angle, some of its velocity is directed upward in the y-direction, and some horizontally in the x-direction.The smaller the angle upward from the ground the more the velocity is in the x-direction and the less in the y-direction. Thus, if the angle is 0 degrees relative to the ground the ball will follow a curved path away from you toward the ground. On the other hand, if the angle is 90 degrees, the ball is going straight up and wll land back in your hand.
To find the x- and y-components of the vector you are given, use sine and cosine functions. Construct a triangle starting from the origin by drawing a hypotenuse at an angle of 25 degrees relative to the ground and label that line 23 m/s. Now at the end of that line drop a line straight down to the x-axis. You now have a right triangle.
The sine of 25 degrees = opposite side/hypotenuse.
You know the angle and the hypotenuse so solve for the opposite side which is the velocity in the y-direction.
sine 25 degrees = 0.4226.
0.4226 = opposite/23 m/s, so opposite side = 9.72 m/s
The cosine of 25 degrees = adjacent side/hypotenuse.
cosine 25 degrees = .9063
.9063 = adjacent side/23 m/s, so adjacent side = 20.845 m/s.
Now you know that the ball is traveling upward at an initial velocity of 9.72 m/s while it is also going horizontally at an initial velocity of 20.845 m/s.
We will assume no resistance from the air, so the ball travels 42 meters at a velocity of 20.845 m/s.
distance = velocity * time, so time = distance/velocity.
time = 42m/20.845m/s = 2.01 s.
In the y-direction, the ball starts at 9.72 m/s but due to the acceleration of gravity downward, slows until its velocity is zero at the highest point.
Using a kinematic equation, we know that Vf = Vi + gt, where Vf is the final velocity, Vi is the initial velocity, g is the acceleration of gravity downward (-9.81 m/s/s), and t is the time in seconds.
So 0 = 9.72 + (-9.81)t
-9.72/-9.81 = .991 s to reach the highest point,and the same time to come back down.
To find the height,we will use another equation.
distance = Vi * t + 1/2 gt^2. Where distance is the height the ball reaches, Vi = 9.72 m/s, t = 0.991s, and g = -9.81m/s/s
distance = 9.72 * 0.991 + 1/2 (-9.81) 0.9991)^2
distance = 4.815 m