# Math Help - Numerical Problem for motion along a plane (HELP)

1. ## Numerical Problem for motion along a plane (HELP)

Question :
A boy kicked a football with an initial velocity component of 15.0m/s and a horizontal velocity component of 22.0m/s.
a)what is the velocity of the football(magnitude and direction)
b)how much time is needed to reach the maximum height?

Attempt:
a)15m/s,N
22.0,E
b) no idea

2. ## Motion of ball

Hello mj.alawami
Originally Posted by mj.alawami
Question :
A boy kicked a football with an initial vertical? velocity component of 15.0m/s and a horizontal velocity component of 22.0m/s.
a)what is the velocity of the football(magnitude and direction)
b)how much time is needed to reach the maximum height?

Attempt:
a)15m/s,N
22.0,E
b) no idea
(a) Assuming that the 15 m/s velocity is the vertical component, the magnitude of the actual velocity is $\sqrt{(15^2 + 22^2)} = 26.63 \,ms^{-1}$ at an angle $\arctan\Big(\frac{15}{22}\Big) = 34.3^o$ above the horizontal.

(b) Again, assuming that the vertical component is initially 15 m/s, the time taken to reach the maximum height is the time taken for a body to come to rest, with an initial velocity 15 m/s and an acceleration $-32 m/s^2$. This is $\frac{15}{32} = 0.47$ sec.

3. Hello, mj.alawami!

A boy kicked a football with an initial vertical velocity component of 15.0 m/s
and a horizontal velocity component of 22.0 m/s.

a) What is the velocity of the football (magnitude and direction)?

The velocity can be represented by this triangle:
Code:
                      *
*   |
*       | 15
* θ         |
* - - - - - - - *
22

The magnitude of the velocity is the length of the hyptenuse.
. . $|v| \:=\:\sqrt{22^2 + 15^2} \:=\:\sqrt{709} \:\approx\:26.6$ m/s.

The direction of the velocity is given by:
. . $\tan\theta \:=\:\frac{15}{22} \quad\Rightarrow\quad \theta \:=\:\arctan\left(\tfrac{15}{22}\right) \:\approx\:34.3^o$

b) How much time is needed to reach the maximum height?

The height $y$ of the football is given by: . $y \:=\:15t - 4.9t^2$

The graph is a down-opening parabola which reaches its maximum at its vertex.
The vertex is at: . $t \:=\:\frac{\text{-}b}{2a} \:=\:\frac{\text{-}15}{2(\text{-}4.9)} \:=\:1.530612245$

The ball reaches maximum height in abut 1.5 seconds.

4. ## Value of g

Hello everyone -

Thanks, Soroban. Yes, of course $g = 9.8$, not $32$. We haven't used $g = 32$ in the UK for aeons, so why I used that value here I don't know.

(Mind you, it's easier to use my method, if not my value: $v = u + at \Rightarrow 0 = 15 - 9.8t \Rightarrow t \approx 1.5$.)