ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: wildareal on October 03, 2010, 12:59:18 am

Title: Motion Question
Post by: wildareal on October 03, 2010, 12:59:18 am: A man dives from a springboard where his centre of gravity is initially 3 m above the water and his initial velocity is 4.9 m/s upwards. Regarding the diver as a particle at his centre of gravity, and assuming that the diver’s motion is vertical, find

a) the diver’s velocity after t seconds
b) the diver’s height above the water after t seconds
c) the maximum height of the diver above the water
d) the time taken for the diver to reach the water.
Title: Re: Motion Question
Post by: moekamo on October 03, 2010, 07:52:09 am: a) $v= u+at = 4.9 - 9.8t$

b) $h= \int 4.9 - 9.8t \ dt = 4.9t - 4.9t^2 + c$

$t=0, \ h=3 \implies c=3 \implies h = 4.9t - 4.9t^2 +3$

c) Max height when $v=0=4.9-9.8t \implies t=0.5 \ \text{seconds}$

So $h=4.9(0.5)-4.9(0.5)^2+3=4.225\ \text{metres}$

d) Reach water $\implies h=0$

so $0= 4.9t - 4.9t^2 +3$

which gives $t=\frac{10}{7}$ or $t=-\frac{3}{7}$

only the postive t value applies, so it takes $\frac{10}{7} \ \text{seconds}$ to reach the water

pretty sure thats all right :)

ATAR Notes | SMF © 2014, Simple Machines