PHYSICS ?????????????

Dimepiece: PHYSICS ?????????????
A cheetah leaves the ground with an initial velocity of 7.0 m/s [up] and an acceleration of 9.81 m/s2 [down]. The cheetah reaches its maximum height in 0.71 s. What is the displacement (height) at the top of the jump? Assume up is positive

i really dont undersatnd, how do u do this, plz giv an explanation, 10 points to best answer 🙂

Answers and Views:

Answer by dan the physics man
y = Vo*t + (1/2)a*t²
y = (7m/s)(0.71s) + (1/2)(9.8 m/s²)(0.71s)²
y = 4.97m + 2.47m
y = 7.44m

hope this helped!

Answer by naveed p
its really easy buddy,
well its intial velocity is 7 but last velocity is zero, i mean at the top of the jump, when it reaches at the top but has not yet started to move downwards, its zero. and acceleration due to gravity is -9.81m/s.
use this equation. i hope u know this equation, cant write it propetly need further explanation ask me. take care
V2 = u2 +2as
U= 7
a= – 9.8
s= let H
V= 0
Gives
(0)2 = (7)2 + 2 (-9.8)H
H= (7 )2/ 2* 9.8
2.5 m

Answer by Naveed T
well its easy,

use H = ut +0.5* a* time square,
well its intial velocity is 7 but last velocity is zero, i mean at the top of the jump, when it reaches at the top but has not yet started to move downwards, its zero. and acceleration due to gravity is 9.81m/s.
use this equation. i hope u know this equation, cant write it propetly need further explanation ask me. take care. the other guy said right but i think he forgot to use time, or he did not see the time is also given.
H= 7*0.71+ 0.5*9.81*0.71
H= 8.38m