Vo = 100m/s[30o].
Yo = 100*sin30 = 50 m/s.
Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g + ho.
y = 0.
g = -9.8 m/s^2.
ho = 100 m.
h = ?
A stone is projected upward at an angle of 30 degree to the horizontal from the top of a tower 100m and it hits the ground at a point q. If the initial veloity of projection is 100ms cal the maximum height of the stone above the groud
5 answers
23
The vertical component of the velocity is needed. Therefore
from v=u + gt
0=usintita +gt
100sin30 = 10t
t=5s.
so it takes 5 sec to get to the max height when thrown from the top of the building. To find the additional distance(height) travel we use this:
S= ut + 1/2gt2.
S= 0 + 1/2*10*25
S= 125m.
Therefore total height is 100+125=225m.
from v=u + gt
0=usintita +gt
100sin30 = 10t
t=5s.
so it takes 5 sec to get to the max height when thrown from the top of the building. To find the additional distance(height) travel we use this:
S= ut + 1/2gt2.
S= 0 + 1/2*10*25
S= 125m.
Therefore total height is 100+125=225m.
I hope that piece above was helpful
How do you solve it
2 answers