Question

at time t=T/4 H/H max is ( where T is the total time of flight h max is the maximum height and H is the height at time t of a projectile launched from the ground)

Answer 1

did any one get this???

Answer 2

I'm still solving it

Answer 3

ok try ur best............

Answer 4

I got 7/16, if that is right answer I'll give you the whole solution.

Answer 5

the answer is 3/4

Answer 6

let's assume that v0y=v0
t1=T/2 time to reach the Hmax height
\[H_{max}=v_{0}t_{1}-gt_{1}^2/2\]
[1]

\[v=v_{0}-gt\]
\[v(t_{1})=0=v_{0}-gt_{1}\]
\[v_{0}=gt_{1}\]
\[g=v_{0}/t_{1}\]
using [1],
\[H_{max}=v_{0}t_{1}-(v_{0}/t_{1})t_{1}^2/2=v_{0}t_{1}/2=v_{0}T/4\]
\[H_{max}=v_{0}T/4\]
\[H(t) = v_{0}t-(v_{0}/t_{1})t^2/2\]
\[H(T/4)=v_{0}T/4-(v_{0}/t_{1})T/4^2/2=v_{0}T/4-(2v_{0}/T)T/4^2/2=\]
\[v_{0}T/4-v_{0}/16= (3/16)v_{0} T\]
\[H=(3/16)v_{0} T\]
now take the ratio

Answer 7

THANXXXXX LAST ONE MORE QUESTION PLZZ SEE IT