A ball is thrown from a point 1.1 m above the ground. The initial velocity is 20.3 m/s at an angle of 32.0° above the horizontal.
(a) Find the maximum height of the ball above the ground???
(b) Calculate the speed of the ball at the highest point in the trajectory???
Vo = 20.3 m/s @ 32 Deg.
Xo = 20.3*cos32 = 17.2 m/s.
Yo = 20.3*sin32 = 10.76 m/s.
a. hmax = 1.1 + (Yf^2-Yo^2/2g,
hmax=1.1 + (0-(10.76)^2) / -19.6=7.0 m.
b. Yf = 0 @ hmax.
To find the maximum height of the ball above the ground (point A) and the speed of the ball at the highest point (point B), we can use the principles of projectile motion.
Let's break down the problem into components:
1. Vertical component:
The ball is thrown upwards, so the gravity will act against its motion. We need to find the time taken by the ball to reach its peak height.
Using the equation vf = vi + at, where:
- vf is the final vertical velocity (0 m/s at the highest point),
- vi is the initial vertical velocity (20.3 m/s * sin(32.0°)),
- a is the acceleration due to gravity (-9.8 m/s^2),
- t is the time taken.
Set vf = 0, substitute the given values, and solve for t:
0 = 20.3 m/s * sin(32.0°) - 9.8 m/s^2 * t
t = (20.3 m/s * sin(32.0°)) / 9.8 m/s^2
Now, to find the maximum height (hmax), use the equation h = vi * t + 0.5 * a * t^2:
hmax = (20.3 m/s * sin(32.0°)) * ((20.3 m/s * sin(32.0°)) / (9.8 m/s^2)) + 0.5 * (-9.8 m/s^2) * ((20.3 m/s * sin(32.0°)) / (9.8 m/s^2))^2
Calculate this expression to find the answer to part (a).
2. Horizontal component:
The horizontal component of velocity remains constant throughout the motion. Therefore, the speed of the ball at the highest point (point B) will be the same as the initial speed.
Calculate the answer to part (b) using the initial velocity: speed at point B = 20.3 m/s.
Remember to use appropriate units and trigonometric functions (e.g., sin, cos) when calculating the values.
Hope this helps!