Just 6 meters high, oh well
downward problem
z = 6 - 9.49 t - 4.9 t^2
upward problem
z = 0 + 9.49 t - 4.9 t^2
so at collision
6 - 9.49 t = 9.49 t
6 = 19 t
t = 0.316 seconds
z = 9.49 (0.316) -4.9 (0.316)^2
= 3 - .5 = 2.5 meters
Two stones are thrown simultaneously. One is thrown straight upward from the base of a cliff and the other is thrown straight downward from the top of the cliff. The height of the cliff is 6 m. The stones are thrown with the same speed of 9.49 m/s. Find the location above the base of the cliff where the stones cross paths.
5 answers
wouldnt the downward problem have +4.9 t^2 because you distribute the minus sign?
Nope, not accelerating upwards
but the upward ball is so i was thinking we would do
downwards distance = 6- (upwards)
= 6 - (9.49t -4.9t^2)
= 6 - 9.49 t + 4.9t^2
downwards distance = 6- (upwards)
= 6 - (9.49t -4.9t^2)
= 6 - 9.49 t + 4.9t^2
V = Vi + a t
h = Hi + Vi t + (1/2) a t^2
here starting from the ground up
Vi is up
a is DOWN (-9.81), in other words it slows down as it rises, and after a while stops rising.
gravity speeds it up coming down, and slows it going up
h = Hi + Vi t + (1/2) a t^2
here starting from the ground up
Vi is up
a is DOWN (-9.81), in other words it slows down as it rises, and after a while stops rising.
gravity speeds it up coming down, and slows it going up