Question

8.) A lady tosses a coin vertically up with a speed of 5.5 m/s. (a) How high will the coin rise before it goes down? (b) How long will the coin stay in air before landing? (c) With what speed will it land on the lady's hand? (d) At what times will the speed of the coin be 3.0 m/s?

Answer 1

Velocity of firing up (u)=5.5 m/sec

(a) $h_{ascent}:$ Maximum height attained by the coin.

$v^2-u^2=2(-g)h..............(g=10 m/sec^2)$

At max. height,v=0

$h=\frac{u^2}{2g}$

$h=\frac{5.5^2}{2 \times 10}$$=1.5125 m$

(b) $t_{air}=t_{ascent}+t_{descent}$

$t_{ascent}=t_{descent}$

$t_{air}=2 \times t_{ascent}$

$t_{ascent}=\sqrt{\frac{2 \times h}{g}}$ $=\sqrt{\frac{2 \times 1.5125}{10}}=0.55sec$

$t_{air}=2 \times 0.55=1.10 m$

(c)

Velocity of firing up=Velocity of landing

so,$v_{landing}=u=5.5 m/sec$

(d)

Velocity of particle is at $3 m/sec$ during two times that is

during ascent and during descent.

1. During ascent:

v=u+at

3=5.5-10t

$t=\frac{2.5}{10}=0.25 sec$

2 During descent:

t will be t_ascent+t1

v=u+at

3=0+10 t1 (at highest point u=0)

$t1=\frac{3}{10}=0.3 sec$

$t=t1+t_{ascent}=0.3+0.55=0.85 sec$