Question

pls help me...
a body thrown vertically up with a speed of 10 m/s from the top of a wall of height 3.45 m. The height from which another body should be dropped simultaneously such that the two bodies reach the ground at the same time is
a. 40
b. 24.5

i got t=u/g =1 s (time up)
h (max) =u^2/2g=100/20=5 m
so total distance down is 5 + 3.45 =8.45
then what to do?? help

Answer 1

What you need here is the total time of flight for the first object. Since it vent up for 1s, as you correctly stated, it'll take 1s to reach the level of the top of the wall again. At that time it's velocity , will again be 10 m/s. The time needed to cover the rest of the way down is:
\[3.45 = h = at^2/2+v*t\]
solution to t is:
\[t = (-v +\sqrt{v^2+4a*3.45})/2a=0.27\]
So the total time of flight is:
1+1+0.27 = 2.27 s
The height of an object, that takes this long to reach the ground is:
\[h = at^2/2 = 10*2.27^2/2=25.76 m\]
so neither is correct, but b is a lot closer.

Answer 2

tnx a lot 4 ur help...i was stuck..

Answer 3

\[2n equals 9\]