3.11 . Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50 s, while Milada jumps horizontally with an initial speed of 95.0 cm/s. How far from the base of the cliff will Milada hit the ground?
SOLUTION
Taking –y to be downward, each cricket moves in projectile motion. As discussed from the book (University Physics 13th Ed, p. 78) both projectiles will have
y-position at any given time, thus we can use the same time $t=3.50s$ for cricket Milada.
Take the following into consideration,
for both
$x_0=y_0=0$ at the origin
$a_x=0$ and $a_y=9.80 / m/s^2$
for Chirpy:
$v_{0x}=0$
$v_{0y}=0$
for Milada:
$v_{0x}=95.0 / cm/s = 0.95 / m/s$
$v_{0y}=0$
Solving for the distance Milada has reached from the base of the cliff,
$x=x_0+v_{0x}t+\frac{1}{2}a_xt^2$
$x=0+v_{0x}t+0$
$ x=$(0.95 / m/s)(3.50s)= 3.32 \ m$
The x and y motion of a projectile are independent with each other and are only related in time in a way that both will reach the ground at the same time.
