A ball is thrown from a cliff upward with a speed of 27 m/s and at an angle of 71 degrees to the horizontal. How high does it go above the cliff to the nearest tenth of a meter?
answer is 33.3.
I am using 27sin71.
That is the initial velocity in the y-direction.
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)+2(-9.8m/s^2)d
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
That is the initial velocity in the y-direction.
Use the following formula:
Vf^2=Vi^2+2gd
Where
d=?
Vi=(27m/s*Sin71º)
Vf=0
g=-9.8m/s^2
Solve for d,
0=(27m/s*Sin 71º)^2+2(-9.8m/s^2)d
0=651.7m/s-19.6m/s^2*d
(-651.7m/s/-19.6m/s^2)=d
d=33.25m=33.3m
To determine the height the ball reaches above the cliff, you need to use the vertical component of the initial velocity.
The given initial velocity has two components: the horizontal component (27 m/s) and the vertical component. To find the vertical component, you need to use trigonometry.
The vertical component of the initial velocity can be determined by multiplying the magnitude of the initial velocity (27 m/s) by the sine of the launch angle (71 degrees).
Therefore, the vertical component of the initial velocity is calculated as:
27 m/s * sin(71 degrees) ≈ 25.3 m/s
This vertical component represents the initial velocity in the upward direction.
Now, to find the height the ball reaches, you can use the formula for vertical displacement:
h = (V₀²)/(2g)
Where:
h = vertical displacement (height)
V₀ = initial vertical velocity (25.3 m/s, rounded to one decimal place)
g = acceleration due to gravity (approximately 9.8 m/s²)
Substituting the given values into the formula:
h = (25.3 m/s)² / (2 * 9.8 m/s²) = 64.09 m (rounded to two decimal places)
Therefore, the ball reaches a height of approximately 64.09 meters above the cliff.