Question

If the ball shown in the figure lands in 2.0 s, about what height was it thrown from?

Answer 1

if it is thrown horizontally, we only need solve the vertical problem
h = Hi - (1/2)gt^2
at ground h = 0
if t = 2 seconds and g = 9.8 m/s^2
0 = Hi - 4.9(2^2)
Hi = initial height = 4*4.9 meters
if working in feet, g = about 32 ft/s^2
and
Hi = 4 * 16 feet

Answer 2

\[y = vt + \frac{ 1 }{ 2 }at ^{2}\]
\[y= (0)t - \frac{ 1 }{ 2 }(9.8m/s ^{2})(2.0s)^{2}\]
\[y = -39.2m\]
so, height is 39.2.

Answer 3

^ typo
y = 19.6m