A third baseman makes a throw to first base 39.7 m away. The ball leaves his hand with a speed of 37.0 m/s at a height of 1.5 m from the ground and making an angle of 19.5 o with the horizontal. How high will the ball be when it gets to first base?
t = 39.7/37cos19.5
y = 37sin19.5 t - 1/2*9.8*t^2 +1.5
To calculate how high the ball will be when it gets to first base, we can use the principles of projectile motion.
First, let's break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component of the velocity can be found using the formula:
vx = v * cosθ
where vx is the horizontal component of the velocity, v is the magnitude of the velocity (37.0 m/s), and θ is the angle of launch (19.5 o).
Plugging in the values, we have:
vx = 37.0 m/s * cos(19.5 o)
Next, let's calculate the vertical component of the velocity using the formula:
vy = v * sinθ
where vy is the vertical component of the velocity.
Plugging in the values, we have:
vy = 37.0 m/s * sin(19.5 o)
Now, we can find the time taken for the ball to reach first base using the horizontal distance and the horizontal component of velocity.
The time taken (t) can be found using the formula:
t = d / vx
where d is the horizontal distance (39.7 m) and vx is the horizontal component of velocity.
Plugging in the values, we have:
t = 39.7 m / vx
Lastly, we can calculate the height of the ball when it gets to first base using the vertical component of velocity, the time taken, and the acceleration due to gravity (assuming the ball is launched from the ground).
The height (h) can be found using the formula:
h = vy * t - (0.5 * g * t^2)
where g is the acceleration due to gravity (9.8 m/s^2).
Plugging in the values, we have:
h = vy * t - (0.5 * g * t^2)
Substituting the values we calculated earlier:
h = (37.0 m/s * sin(19.5 o)) * (39.7 m / vx) - (0.5 * 9.8 m/s^2 * (39.7 m / vx)^2)
Now, we can solve for h by substituting the values and calculating the expression.