A ball is thrown vertically upwards with an initial velocity of 20.96 m/s. Neglecting air resistance, how long is the ball in the air? What is the greatest height reached by the ball?
this problem seems easy enough but I can't seem to get the answer. How can I get time using only velocity?
Two equations for constant acceleration:
v = Vo + a t
x = Xo + Vo t +(1/2) a t^2
here
a = -9.8 m/s^2
Vo = 20.96
Xo = 0
when v = 0, we are at the top
solve for t at the top (t at the ground again will be twice that by the way)
Then use that t at the top in the second equation
x = 0 + 20.96 t -4.9 t^2
to get the height at the top
if you did not belive me about doubling that t at the top to get t when it hits the ground again, there are two ways to proceed.
Way 1
let the ball drop with zero initial velocity from the top (you will see why the time falling is the same as the time rising)
Way 2
solve the second equation for t when x= 0,
of course t = 0 and t = something which I hope is t = twice t at the top should be solutions to the quadratic.
To find the time the ball is in the air, you can use the formula:
t = (2 * v) / g
Where:
t is the time in seconds
v is the initial velocity in meters per second
g is the acceleration due to gravity, which is approximately 9.8 m/s²
In this case, the initial velocity of the ball is given as 20.96 m/s. So, you can substitute the values into the formula:
t = (2 * 20.96) / 9.8
t ≈ 4.276 seconds
Therefore, the ball is in the air for approximately 4.276 seconds.
Now, let's find the greatest height reached by the ball. To do this, we can use the equation:
h = (v^2) / (2 * g)
Where:
h is the maximum height reached by the ball in meters
v is the initial velocity in meters per second
g is the acceleration due to gravity, which is approximately 9.8 m/s²
Substituting the given values:
h = (20.96^2) / (2 * 9.8)
h ≈ 22.26 meters
Therefore, the greatest height reached by the ball is approximately 22.26 meters.
To find the time the ball is in the air, you can use the equation of motion for vertically thrown objects:
v = u + at
where:
v is the final velocity (in this case, 0 m/s, since the ball reaches its maximum height and then starts falling down),
u is the initial velocity (20.96 m/s),
a is the acceleration (due to gravity, which is approximately 9.8 m/s²),
and t is the time.
Since the ball reaches its highest point when its final velocity is 0, we can replace the final velocity in the equation with 0:
0 = 20.96 - 9.8t
Rearranging the equation, we get:
9.8t = 20.96
Dividing both sides by 9.8:
t = 20.96 / 9.8
Calculating this, we find:
t ≈ 2.1388 seconds
Therefore, the ball is in the air for approximately 2.1388 seconds.
To find the greatest height reached by the ball, we can use the equation:
s = ut + (1/2)at²
where:
s is the displacement (change in height),
u is the initial velocity (20.96 m/s),
a is the acceleration (due to gravity, -9.8 m/s²),
and t is the time (2.1388 seconds).
Substituting the values into the equation, we get:
s = (20.96)(2.1388) + (1/2)(-9.8)(2.1388)²
Calculating this expression, we find:
s ≈ 22.152 meters
Therefore, the greatest height reached by the ball is approximately 22.152 meters.