A boy kicks a soccer ball directly at a wall 41.8 m away. The ball leaves the ground at 42.7 m/s with an angle of 33.0 degrees to the ground. What height will the ball strike the ground?
distance = 41.8 m
theta = 33.0 degrees
Vi = 42.7 m/s
I really really need your help. pretty pleaseee.. thank you :D KAMSAHAMNIDA
1. Resolve the initial velocity.
2. Find the time taken for the horizontal velocity of the ball to hit the wall.
3. Using the value of the time found in 2,find the height reached by the vertical velocity of the ball.
is the distance = 27.60 m
time = 1.17s
KAMSAHAMNIDA
To find the height at which the ball will strike the ground, we need to break down the initial velocity into horizontal and vertical components.
The horizontal component of the velocity (Vx) remains constant throughout the motion.
Vx = Vi * cos(theta)
Here, Vi is the initial velocity and theta is the angle of projection. Let's plug in the values:
Vx = 42.7 m/s * cos(33.0 degrees)
Vx = 42.7 m/s * 0.8387
Vx ≈ 35.8149 m/s
Now, let's find the time it takes for the horizontal motion of the ball using the equation:
distance = velocity * time
41.8 m = 35.8149 m/s * time
Solving for time:
time ≈ 1.1672 seconds
Next, we'll find the initial vertical velocity component (Vy):
Vy = Vi * sin(theta)
Vy = 42.7 m/s * sin(33.0 degrees)
Vy = 42.7 m/s * 0.5446
Vy ≈ 23.2302 m/s
Now, we'll use the equation for vertical motion:
height = Vy * time + 0.5 * acceleration * time^2
Since the ball is moving vertically against gravity, the acceleration is -9.8 m/s^2 (negative because it is acting in the opposite direction).
height = 23.2302 m/s * 1.1672 s + 0.5 * (-9.8 m/s^2) * (1.1672 s)^2
Simplifying:
height ≈ 27.0659 m
Therefore, the ball will strike the ground at a height of approximately 27.0659 meters.