1. What is the maximum height that a basketball will have if thrown up into the air from a height of six feet, with an initial speed of 20 feet per second? (Must use quadratic formula).

2. After how many seconds does the basketball hit the ground? (Must solve by completing the square)

Question

1.  What is the maximum height that a basketball will have if thrown up into the air from a height of six feet, with an initial speed of 20 feet per second?  (Must use quadratic formula).



 2.  After how many seconds does the basketball hit the ground?  (Must solve by completing the square)

rvc · Answer

Initial velocity,u=20m/s

rvc · Answer

considering upwards positive 
then acceleration due to gravity is negative

rvc · Answer

|dw:1448204496277:dw|

rvc · Answer

considering 
v=u+at

since at h_max v=0
that is final velocity will be 0
therefore
u will get time...
as u=20feet/sec
g=9.81 m/s
there has to be a conversion ....

rvc · Answer

or else

v^2=u^2+2 a s

rvc · Answer

well @Joe-Palooka do you have any idea?

rvc · Answer

like u might have solved similar question

rvc · Answer

its a simple physics kinematics problem :/

anonymous · Answer

Hi.  Thanks, but I am still lost, since I need to use the quadratic equation on the first part and then use "completing the square" for part two.  I think I have part one done.

rvc · Answer

please show us the working for part 1

rvc · Answer

@Astrophysics

astrophysics · Answer

$$v_0 = 20 ft/s,~t=0,~h_0= 6 ft$$
$$y(t) = -g t^2+v_0t+h_0$$ this is your equation for a free falling body

astrophysics · Answer

Or you may use $$y(t) = -16t^2+v_0t+h_0$$ where that g is in ft/s^2

astrophysics · Answer

To find the max height you can just use the vertex, but understand what it is $$h = \frac{ -b }{ 2a }$$

anonymous · Answer

Thank you all.  Very helpful.

rvc · Answer

my pleasure...
All the best :)

anonymous · Answer

g=-32 ft/sec
u=20 ft/s
v=0 ft /s
$$v=u+\gt,0=20-32t,t=\frac{ 20 }{ 32 }=\frac{ 5 }{ 8 } \sec,h=ut+\frac{ 1 }{ 2 }g t^2+6$$
$$h=20 \times \frac{ 5 }{ 8 }+\frac{ 1 }{ 2 }\times \left( -32 \right)\left( \frac{ 5 }{ 8 } \right)^2+6$$
=?

anonymous · Answer

correction v=u+ gt not v=u+>

mrnood · Answer

the direct solution is obtained from the equation

v^2 = u^2 +2as

v is velocity at peak (=0)
u is intitial velocity = 20 ft/s
a = g = -32ft/s^2

s is your solution - then add the intial 6 ft

cheesecakekitten · Answer

http://openstudy.com/study#/updates/5651f61de4b0aa790266f390

jagr2713 · Answer

Great job guys ! Thanks for helping