HELP ASAP//BRAINLEST+THANKS: Subj:gravity formula stuff A rock is thrown upward with a velocity of 21 meters per second from the top of a 48 meter high cliff, and it misses the cliff on the way back down. When will the rock be 7 meters from ground level? Round your answer to two decimal places.
seconds..
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Use Gravity Formula:
h=−12gt^2+v0t+h0
where g=32ft/s^2= 9.8 m/s^2

6.46

Step-by-step explanation:

h(t) = (1/2) + v0t + h0

Origin is at surface of water (h = 0), up is positive.

a = -g = -9.8 m/

v0 = 25 m/s

h0 = 50 m

h(t) = -4.9t2 + 25t + 50

Find t when

7 = + 25t + 50

- 25t - 43 = 0

t = [1/(2*4.9)] [25 ±√( + 4*4.9*43)]

t = (0.102)(25 ± 38.312)

t > 0 ⇒ use only the positive square root

t = (0.102)(25 + 38.312) ≅ 6.46 s

It is 7m above the water at about 6.46 s

the answer is a i belaeve atleast thats what my calculator just gfave me

step-by-step explanation: