Student A is on a 9m Hugh bridge while Student B is on the ground. Student B throws an apple up to Student B with a velocity of 18m/s. How long does it take for the apple to reach student A's hand? if student A is unable to catch the apple, how long would it take for the apple to come back to the ground?
2 answers
sorry I meant 9m high bridge
a. h = Vo*t + 0.5g*t^2 = 9
18t - 4.9t^2 = 9
-4.9t^2 + 18t - 9 = 0
t=0.597 s. and 3.08 s(Use quad. formula)
Choose t = 0.597 s since the apple reaches its max ht. of 16.5 m, in 1.84 s.
V = Vo + g*t = 0 @ max ht.
18 - 9.8*t = 0
9.8t = 18
Tr = 1.84 s. = Rise time.
b. If student A does not touch the ball it will reach a max ht. of 16.5 m.:
h = 18*1.84 - 4.9*1.84^2 = 16.5 m, max.
Tf = Tr = 1.84 s. = Fall time.
Tr+Tf = 1.84 + 1.84 = 3.68 s. = Time to
return to gnd.
18t - 4.9t^2 = 9
-4.9t^2 + 18t - 9 = 0
t=0.597 s. and 3.08 s(Use quad. formula)
Choose t = 0.597 s since the apple reaches its max ht. of 16.5 m, in 1.84 s.
V = Vo + g*t = 0 @ max ht.
18 - 9.8*t = 0
9.8t = 18
Tr = 1.84 s. = Rise time.
b. If student A does not touch the ball it will reach a max ht. of 16.5 m.:
h = 18*1.84 - 4.9*1.84^2 = 16.5 m, max.
Tf = Tr = 1.84 s. = Fall time.
Tr+Tf = 1.84 + 1.84 = 3.68 s. = Time to
return to gnd.
1 answer