You throw a ball with a speed of 25 m/s at an angle of 40.0° above the horizontal directly toward a wall (Fig. 4-35). The wall is 22.0 m from the release point of the ball. a)How long does the ball take to reach the wall? b)How far above the release point does the ball hit the wall? c)What are the horizontal and vertical components of its velocity as it hits the wall?
I will be happy to critique your work.
You can figure the horizontal component of velocity, you have distance horizontal..figure time from that.
For the final altitude, use vertical distance
y= Vovertical*time -1/2 g time^2
a) To find out how long the ball takes to reach the wall, we can use the horizontal motion of the ball. The given information includes the initial speed of the ball (25 m/s) and the distance to the wall (22.0 m). We can use the equation for horizontal distance:
Horizontal distance = initial horizontal speed * time
Since there is no acceleration in the horizontal direction, the initial horizontal speed remains constant throughout the motion. Therefore, the equation reduces to:
Distance to the wall = initial horizontal speed * time
Rearranging the equation to solve for time:
time = Distance to the wall / initial horizontal speed
Substituting the given values:
time = 22.0 m / 25 m/s
Using a calculator, we can find that the time is approximately 0.88 seconds.
b) To find the vertical distance above the release point where the ball hits the wall, we can use the equation for vertical motion. The given information includes the initial speed of the ball (25 m/s) and the launch angle (40.0°). The vertical distance can be calculated using the equation:
Vertical distance = initial vertical speed * time - (1/2) * gravity * time^2
The initial vertical speed can be found by decomposing the initial speed into its vertical and horizontal components:
Initial vertical speed = initial speed * sine(angle)
Substituting the given values:
Initial vertical speed = 25 m/s * sin(40.0°)
Using a calculator, we find that the initial vertical speed is approximately 16.94 m/s.
Now, we can substitute the values into the equation for vertical distance:
Vertical distance = (16.94 m/s) * (0.88 s) - (1/2) * (9.8 m/s^2) * (0.88 s)^2
Evaluating this expression, we find that the vertical distance is approximately 9.73 meters.
c) To determine the horizontal and vertical components of the velocity as it hits the wall, we can use the fact that the horizontal velocity remains constant while the vertical velocity changes due to gravity.
The horizontal component of velocity remains the same throughout the motion and is equal to the initial horizontal speed, which is 25 m/s.
The vertical component of velocity at the moment the ball hits the wall can be determined using the equation:
Final vertical velocity = initial vertical velocity - gravity * time
Substituting the given values:
Final vertical velocity = (16.94 m/s) - (9.8 m/s^2) * (0.88 s)
Evaluating this expression, we find that the final vertical velocity is approximately 8.01 m/s.
Therefore, the horizontal component of velocity as it hits the wall is 25 m/s, and the vertical component of velocity is 8.01 m/s (upward direction).