Question

A ball is projected upward at time t = 0.0 s. from a point on a roof 30 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 24.5 m/s. Consider all quantities as positive in the upward direction.

Answer 1

wat u hav to find??

Answer 2

@tpond23

Answer 3

At time t = 2.50 s, the acceleration of the ball, in m/s2, is closest to

At time t = 2.50 s, the acceleration of the ball, in m/s2, is closest to
+5
zero
-10
+10
-5

Answer 4

At time t = 2.50 s, the velocity of the ball, in m/s, is closest to

At time t = 2.50 s, the velocity of the ball, in m/s, is closest to
+24
zero
-17
-24
+17

Answer 5

The average velocity (in m/s) of the ball, during the first 4.0 s, is closest to

The average velocity (in m/s) of the ball, during the first 4.0 s, is closest to
-5
zero
-10
+10
+5

Answer 6

The velocity of the ball, in m/s, when it is 24 m above the ground is closest to


The velocity of the ball, in m/s, when it is 24 m above the ground is closest to
-11
-16
-27
-22
-33

Answer 7

The time, in seconds, when the ball strikes the ground is closest to

The time, in seconds, when the ball strikes the ground is closest to
6.2
6.4
5.6
5.8
6.0

Answer 8

Problem has only 1 direction, up and down, so there is no need for me to resolve forces into their horizontal and vertical components.

Equations that I might need:
\[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}\:\:\:(1)\]\[v=v_{0}+at\:\:\:(2)\]\[v^{2}=v_{0}^{2}+2a(x-x_{0})\:\:\:(3)\]\[x-x_{0}=\frac{v_{0}+v}{2}t\:\:\:(4)\]Where v0, x0 means the initial speed.

I take upwards as positive.

"At time t = 2.50 s, the acceleration of the ball, in m/s2, is closest to"
I will take the answer as -10m/s^2 because I have assumed that the experiment is done on Earth, and it doesn't really matter if the ball went 10m up or 10m down. It's still roughly the same at 9.8m/s^2. And I have taken upwards as positive, so acceleration of the ball is -10 (because downwards). Note that the ball's acceleration is ALWAYS downwards due to gravity. So what happens is the ball slows down after being thrown up, slow to 0m/s, then goes in the opposite direction (that is down!).

"At time t = 2.50 s, the velocity of the ball, in m/s, is closest to"
I have x0=+30m, a=-9.8 and v0=+24.5m/s and t=2.5s. I want to find speed at a particular moment in time. Equation 1 finds me distance at 2.5 seconds, but that's not I want. Equation 2 finds me velocity at 2.5 second. That's what I want!
\[v=v_{0}+at=24.5+(-9.8)(2.5)=m/s=0\] Even though the speed is zero, the acceleration of the ball is still -9.8m/s because F=ma and F is the weight of the ball.

"The average velocity (in m/s) of the ball, during the first 4.0 s, is closest to"
Average velocity is the total displacement divided by time. So i use equation (1) to find the distance the ball is at in t=4s. \[x=x_{0}+v_{0}t+\frac{1}{2}at^{2}=30+24.5*4+\frac{1}{2}(-9.8)(4)^2=49.6m\]
So the total displacement is
\[49.6-30=19.6m\:(positive\:so\:upwards)\]
and thus average velocity is
\[19.6/4=4.9ms^{-2}\]

"The velocity of the ball, in m/s, when it is 24 m above the ground is closest to"
For this part I use equation (3)
\[v^{2}=v_{0}^{2}+2a(x-x_{0})\]Subbing in the values for v0=+24.5, a=-9.8 and x-x0= 24-30.
solving for v, I get v=-26.8m/s (note that are 2 answers because I take square root, and you need to think which answer makes sense. The ball is at 24m, which means it's below the roof, this only happens after 2.5s, which means that the ball's velocity is negative or downwards)
I could've used equation (1) and (2) but it would involve me solving quadratic equation for t.

"The time, in seconds, when the ball strikes the ground is closest to"
When the ball touches the floor, v is not 0 yet! It is only after collision that the velocity could be zero, but that's another thing altogether.
Using equation (3):
\[v^{2}=24.5^2+2(-9.8)(0-(+30))=1188.25\]
\[v=-34.5m/s\]
Then I use equation (4) to find t, which is 6.02s

Answer 9

thank you so much @Festinger for your awesome help and great answers as always!