SOLUTION: A ball is thrown off a 30 m cliff at 20 m/s at an angle of 60 degrees above the horizontal
What is the maximum height of the object?
What is the total hang time of the ball?
Wh
Algebra.Com
Question 1170979: A ball is thrown off a 30 m cliff at 20 m/s at an angle of 60 degrees above the horizontal
What is the maximum height of the object?
What is the total hang time of the ball?
What is the final velocity of the ball?
Found 2 solutions by htmentor, Solver92311:
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The total height will be h0 + h, where h0 is the height of the cliff
We can use the formula v2^2 = v1^2 + 2ah, with v1 = vy = 20*sin(60), and a = g
h = vy^2/2g
I assume total hang time refers to the time the ball takes to hit the ground.
The time to reach maximum height is determined from v2 = v1 + at,
with v2 = 0. Thus t = vy/g
The time to fall from the maximum height to the ground is determined from
h + h0 = v1t + 1/2at^2. Since v1 = 0, we have t = sqrt(2(h + h0)/g)
Now for the final velocity, we need to determine the x and y components
vx will remain constant throughout, and is equal to 20*cos(60)
The y component of the final velocity is obtained from v2^2 = v1^2 + 2a(h+h0)
with v1 = 0
I will leave it to you to fill in the values and get the final numbers
Answer by Solver92311(821) (Show Source): You can put this solution on YOUR website!
Since you provided no information on temperature, precipitation, air density, wind speed, or wind direction, I have to presume that this experiment was conducted in some sort of huge evacuated container and the effects of the atmosphere on the projectile are not to be considered.
The height in meters of a projectile near the surface of the earth at 
seconds after it is launched is given by:
 \ \ \ \ \ \ \ \ h(t) =\ -4.9t^2\ +\ \sin(\theta)v_ot\ =\ h_o)
Where 
is the initial velocity in meters per second of the projectile, 
is the angle of launch, and 
is the initial height in meters. Then v_o)
is the vertical component of the initial velocity.
The instantaneous vertical velocity is given by the first derivative of the height function:
 \ \ \ \ \ \ \ \ \frac{dh}{dt}\ =\ v(t)\ =\ -9.8t\ +\ \sin(\theta)v_o)
The maximum height of the projectile is the value of the height function at the time the velocity function is zero. So set equation (2) equal to zero and solve for 
and then calculate the value of )
.
The projectile will be in the air for the amount of time it takes for the ball to reach zero height. Set the height function equal to zero and then solve for the positive root which will be 
.
The final velocity will be the value of )
.
UPDATE
The final vertical velocity will be the value of .
The horizontal velocity is a constant v_o)
The final velocity magnitude is ^2\,+\,\(\cos(\theta\)v_o\)^2})
The final velocity direction is }{\(\cos(\theta\)v_o}\))
Make sure your calculator is set to degrees when you do the arctan calculation.
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
From
I > Ø
RELATED QUESTIONS
A long-jumper leaves the ground at an angle of 20º above the horizontal at a speed of 11 (answered by ikleyn)
A ball is given an initial velocity of 10.0 m/s at 37° above the horizontal. Assume that (answered by ikleyn)
An artillery shell is fired with an initial velocity of 100 m/s at an angle of 30 degrees (answered by Alan3354)
(d) A ball is thrown vertically upwards from ground level with a velocity of 28 m/s
(i)... (answered by ankor@dixie-net.com)
A ball is thrown off the top of building. At time t (in seconds), its height (in feet)... (answered by ikleyn)
A stone is thrown horizontally at a speed of +5.0m/s off the top of a cliff 78.4m high.
(answered by ikleyn)
A football is kicked at an angle of 37 degrees with a velocity of 20 m/s. Calculate:
a.) (answered by Alan3354)
A ball is thrown across a playing field from a height of 5 ft above the ground at an... (answered by MathLover1)
Mae is on top of the building and Jea is down. If Mae throws a ball at an angle of 60... (answered by Edwin McCravy)