no initial vertical speed
so
v = 0 -9.8 t
v at water = -9.8*1.8 = - 17.6 m/s
h = Hi + Vi t -4.9 t^2
0 = Hi + 0 -4.9(1.8)^2
Hi = 4.9* 1.8^2 = 15.9 m
(very high indeed)
The next part of the question is of course how far did Janet land from under the platform:
d = U t = 1.73 (1.8) = 3.11 m
Janet jumps off a high diving platform with a horizontal velocity of 1.73 m/s and lands in the water 1.8 s later.
How high is the platform? The acceleration of gravity is 9.8 m/s2 .
Answer in units of m
2 answers
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