A ball is thrown from the window of a high-rise building with an initial velocity of 7.7 m/s at an angle of 34 deg. below the horizontal. The bill strikes the ground 4.2 seconds after being thrown. Find the height from which the ball was thrown.
y = H -7.7 sin34 t - 4.9 t^2.
y = 0 at t = 4.4 s
Substitute those y and t values into the first equation and solve for the initial height H
To find the height from which the ball was thrown, we can use the equations of motion. The key is to break down the motion of the ball into horizontal and vertical components.
First, let's analyze the horizontal component of the motion. The initial velocity of the ball is given as 7.7 m/s, and the angle of 34 degrees below the horizontal does not affect the horizontal velocity. Therefore, the horizontal component of the velocity remains constant throughout the motion. We can use the formula:
Horizontal distance = horizontal velocity × time
Since the horizontal component of the velocity remains constant at 7.7 m/s, we have:
Horizontal distance = 7.7 m/s × 4.2 s
Horizontal distance = 32.34 m
Now, let's analyze the vertical component of the motion. The initial velocity of the ball has two components: one along the x-axis (horizontal), and the other along the y-axis (vertical). The initial vertical velocity can be found using trigonometry:
Initial vertical velocity = initial velocity × sin(angle)
Initial vertical velocity = 7.7 m/s × sin(34°)
Initial vertical velocity = 4.125 m/s
We can use the formula for vertical displacement in projectile motion:
Vertical displacement = (initial vertical velocity × time) + (0.5 × acceleration × time^2)
Since the ball strikes the ground, the vertical displacement is equal to the initial height:
Initial height = (initial vertical velocity × time) + (0.5 × acceleration × time^2)
The acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in the values we have:
Initial height = (4.125 m/s × 4.2 s) + (0.5 × 9.8 m/s^2 × (4.2 s)^2)
Initial height = 17.325 m + 87.156 m
Initial height = 104.481 m
Therefore, the height from which the ball was thrown is approximately 104.481 meters.