A body is projected with a velocity of 200m/s at an angle of 30 above the horizontal. Take g to be 10m/s2 calculate the time taken for the projectile to reach the maximum height.
4 answers
10 s
Rubbish
Vi = initial speed up = 200 sin 30 = 100 m/s
v = speed up = Vi - g t
at top v = 0
0 = 100 - 10 t
t = 10 seconds to top
v = speed up = Vi - g t
at top v = 0
0 = 100 - 10 t
t = 10 seconds to top
by the way the max height Ht results easily
h = 0 + Vi t - (g/2) t^2
so
Htop = 100*10 - 5*100 = 1000-500 = 500 meters
also it takes10 sec upward, another ten down so in air for 20 s
u = horizontal speed = 200 cos 30
so range = 200 cos 30 * 20 = 3464 meters
h = 0 + Vi t - (g/2) t^2
so
Htop = 100*10 - 5*100 = 1000-500 = 500 meters
also it takes10 sec upward, another ten down so in air for 20 s
u = horizontal speed = 200 cos 30
so range = 200 cos 30 * 20 = 3464 meters
1 answer