In 1978, Geoff Capes of the United Kingdom won a competition for throwing 5 lb bricks; he threw one brick a distance of 44.0 m. Suppose the brick left Capes' hand at an angle of 45.0 degrees with respect to the horizontal.
A. What was the initial speed of the brick?
i got 20.8 m/s.
B. What was the maximum height reached by the brick?
i got 14.7 m.
C. If Capes threw the brick straight up with the speed found in (a), what would be the maximum height the brick could achieve?
i don't get C.
PLEASE HELP!!! THANK YOU.
2007-10-04 09:07:30 · 1 answers · asked by Anonymous in Education & Reference ➔ Homework Help
v^2 = u^2 + 2ax
(^2 means squared)
u = initial velocity (20.8m/s, according to you)
v = final velocity, in this case 0, when the brick is at full height
a = acceleration, 9.8m/s towards the ground
x = height the brick got to.
so, 2ax = v^2 - u^2
x = (v^2 - u^2) / 2a
2007-10-04 09:13:05 · answer #1 · answered by Yanni Depp 6 · 0⤊ 0⤋