A 1.5 tall student throws a baseball with a horizontal component of velocity of 25m/s. It takes 3.00s to come back to its original height.
d. how far did the ball travel horizontally?
e. What is the initial velocity in the y-direction?
f.At what angle is the baseball thrown?
d. d = Xo*T = 25m/s * 3s. = 75 m.
e. Tr = 3s/2 = 1.50 s. = Rise time.
e. Y = Yo + g*Tr = 0.
Yo - 9.8*1.5 = 0.
Yo = 14.7 m/s = Ver. component of initial velocity.
f. Tan A = Yo/Xo = 14.7/25 = 0.58800.
A = 30.5o.
To find the answers to these questions, we can use the kinematic equations of motion and apply them to the given information.
First, let's break down the given information:
- Height of the student (h) = 1.5 m
- Horizontal component of velocity (Vx) = 25 m/s
- Time taken to reach the original height (t) = 3.00 s
Now let's proceed to find the answers to each question step by step:
d. How far did the ball travel horizontally?
To find the horizontal displacement (distance), we need to calculate the total time the ball spends in the air. Since the time taken to reach the original height is the time for one complete cycle, the total time in the air will be twice that.
Total time in the air (T) = 2 * t = 2 * 3.00 s = 6.00 s
Now we can calculate the horizontal distance (d) traveled by the ball using the equation:
d = Vx * T
d = 25 m/s * 6.00 s
d = 150 m
Therefore, the ball traveled a horizontal distance of 150 meters.
e. What is the initial velocity in the y-direction?
To find the initial velocity in the vertical (y) direction, we need to consider the equation for vertical motion. Since the ball reaches its original height after 3.00 seconds, we can use this time and the known height to find the initial vertical velocity (Vy).
The equation for vertical motion is:
h = Vyi * t + (1/2) * g * t^2
Where:
h = vertical displacement (1.5 m)
Vyi = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2, considering upward direction as positive)
t = time (3.00 s)
Plugging in the values:
1.5 m = Vyi * 3.00 s + (1/2) * (-9.8 m/s^2) * (3.00 s)^2
Simplifying the equation:
1.5 m = 3 Vyi - 44.1 m
Rearranging the equation:
3 Vyi = 45.6 m
Vyi = 45.6 m / 3
Vyi = 15.2 m/s
Therefore, the initial velocity in the y-direction is 15.2 m/s.
f. At what angle is the baseball thrown?
To find the angle at which the baseball is thrown, we can use the formula for the projectile motion's launch angle (θ):
θ = arctan(Vyi / Vx)
Plugging in the values we found:
θ = arctan(15.2 m/s / 25 m/s)
Using a calculator or trigonometric table, we find:
θ ≈ 31.78 degrees
Therefore, the baseball was thrown at an angle of approximately 31.78 degrees.