Question

A ball is thrown straight upward and returns
to the thrower’s hand after 4.2 s in the air. A
second ball is thrown at an angle of 38◦ with
the horizontal.

At what speed must the second ball be
thrown so that it reaches the same maximum
height as the first? The acceleration of gravity
is 9.8 m/s^2.

Answer in units of m/s

Answer 1

@thewinterfawn

Answer 2

um.. I don't know... that's why I'm asking. ^-^

Answer 3

I think it might be 0.

Answer 4

Can you show your work?

Answer 5

Oh.. okay. thanks (:

Answer 6

That's not all!

Answer 7

It's not? O.o

Answer 8

How do you find the speed?

Answer 9

Do you still need help?

Answer 10

Yes please. (:

Answer 11

Well. The first ball cannot be thrown with 0 initial velocity. The ball would fall to the floor in that case.

Answer 12

So, it is needed to find the maximum height of the first ball, but knowing that there should be some initial velocity.

Answer 13

Do you understand that?

Answer 14

Well, to find the initial velocity of the first ball, we use this equation,
\[v=v_0-gt\]The ball needs 4.2 s to come back. So the maximum height, the place where the ball has v=0, it reached at 4.2/2=2.1 s. So,
\[0=v_0-9.8\cdot2.1\Rightarrow v_0=20.58\ m/s\]

Answer 15

Now, we know the initial velocity of the first ball. So we can find the height that the first ball can reach,
\[h=h_0+v_0t-\frac{1}{2}gt^2\]The maximum height is reached at 2.1 s, so,
\[h=0+20.58\cdot2.1-4.9\cdot (2.1)^2=21.61\ m\]

Answer 16

Now, we know the maximum height. We can apply that, when the throw is made with an angle,
\[v_{0x}=v_0\cos\theta\]\[v_{0y}=v_0\sin\theta\]\[y=y_0+v_{0y}t-\frac{1}{2}gt^2\]
But we need to find the time when the ball reach that height,
\[v_y=v_{0y}-gt\]

Answer 17

We apply our values,
\[21.61=v_{0}\sin(38)t-4.9t^2\] And
\[0=v_0\sin(38)-9.8t\Rightarrow t=v_0\sin(38)/9.8=0.0628\cdot v_0\] So, using the first equation,
\[21.61=0.0386775v_0^2-0.0193248v_0^2\]

Answer 18

Solving,
\[v_0=33.42\ m/s\]

Answer 19

That is the final answer ;).

Answer 20

Do you have the final answer to check it?

Answer 21

Oops sorry I was doing my calculus homework. Thank you!

Answer 22

No problem, I post the answer in various replies, because the navigator crash eventually :).