Question

a ball is thrown vertically upward from the ground with an initial speed of 25 m/s; at the same instant another ball is dropped from a building 15 m high. after how long will the balls be at the same height? what are their speeds when they are at the same height?

Answer 1

object 1
\[\Large X_{1_F} - X_{1_i} = V_{1_i}t+ \frac 12 a_{1}t^2\]
\[\Large \implies X_{1_F} = V_{1_i}t + X_{1_i}+ \frac 12 a_{1}t^2\]
object 2
\[\Large \implies X_{2_F} - X_{2_i} = V_{2_i}t + \frac 12 a_{2}t^2\]
\[\Large \implies X_{2_F} = V_{2_i}t + - X_{2_i} +\frac 12 a_{2}t^2\]
where X is the distance
i - initial
f = final
t = time
V = velocity

do you agree with that?

Answer 2

yes.

Answer 3

an ideal way is to draw a diagram first..

Answer 4

so when they meet then the X_1F and X_2F would be equal

Answer 5

since final distance of object 1 = final distance of object 2

Answer 6

\[\large V_{1_i} t + X_{1_i} + \frac 12 a_1 t^2 = V_{2_i} t + X_{2_i} + \frac 12 a_2 t^2\]
right?

Answer 7

right good omni