Posted by Sahil Patel on Wednesday, September 7, 2011 at 6:42pm.
Vf^2 = Vo^2 + 2gd,
d = (Vf^2 - Vo^2) / 2g,
d = 7 + (0 - (54)^2) / -64 = 52.56 Ft.
= Distance above ground.
Vf = Vo + gt,
t(up) = (Vf - Vo) / g,
t(up) = (0 - 54) / -32 = 1.69s.
d = Vo*t + 0.5gt^2 = 52.56 Ft.
0 + 0.5*32t = 52.56,
16t = 52.56,
t(down) = 3.29s.
T = t(up) + t(down) = 1.69 + 3.29 = 4.98s = Time to reach ground.
Correction:
0 + 0.5*32t^2 = 52.56,
16t^2 = 52.56,
t^2 = 3.29,
t(down) = 1.81s.
T = t(up) + t(down) = 1.69 + 1.81 = 3.50s. = Time to reach ground.
Related Questions
Alegra 2 - A baseball is hit at a point of 3 feet above ground at a velocity of ...
Physics - The velocity of the baseball as it leaves the bat is 38.2 m/s. ...
Math - A baseball is thrown up in the air from a height of 3 feet with an ...
Precalc - Fred hits a baseball 3 ft above ground at a velocity of 100 ft per sec...
Physics - A baseball is thrown with an initial velocity of 37 m/s at an angle of...
Algebra 2 - A baseball player swings and hits a pop fly straight up in the air ...
Physics - Baseball player A bunts the ball by hitting it in such a way that it ...
Physics! - Baseball player A bunts the ball by hitting it in such a way that it ...
physics - A baseball player hits the ball at a speed of 20.1 m/s and an angle of...
Physics - Baseball player A bunts the ball by hitting it in such a way that it ...
For Further Reading
