finding the maximum height of an arrow shot in the air

[christinaa_s]

QUESTION: An archer shoots an arrow straight up in the air with an initial velocity of 45 m/s [U], it takes the arrow 4.6 s to reach its maximum height. Calculate this maximum height.

the α, ∆d, and vi should have arrows over them:

MY ANSWER:
displacement = ½ [(α ∙ ∆t + vi) + vi + vi] ∆t
therefore vi ∆t + ½ α t(2)
therefore ∆d = 207 + 103.68
therefore ∆d = 310.68 m

I'm not sure if i used the correct formula or if my answer if right.
Also, how do you draw vector diagrams to scale? :shy:

[Kurdt]

Remember that acceleration due to gravity is -g. Other than that all the other umbers seem fine apart from the error leading to the final answer.

[christinaa_s]

another question i am not sure of!..

QUESTION: A bird takes 8.5 s to fly from position A (va= 4.4 m/s (3 ° S of E) to position B(vb= 7.8 m/s ( 25° N of E) along the path. Find the bird's average acceleration.

MY ANSWER:
vax = 4.4 cos 31
vay = -4.4 sin 31

vbx = -7.8 cos 25
vby = 7.8 sin 25

ax = vbx – vax / t therefore ax = -7.8 cos 25 - 4.4 cos 31 / 8.5 ?
ay = vby – vay / t therefore ay = 7.8 sin 25 + 4.4 sin 31 / 8.6 ?

therefore a = √ ax(2) + ay(2)
a = √ 1.5 + 1.06
a= 1.60 m/s ?

once again i am not sure if my answer is correct, and if vay/ vbx should be negative..

[GregA]

Also, how do you draw vector diagrams to scale? :shy:

Imagine 2 forces..one has a magnitude of 4N and acts along the x axis, the other has a magnitude of 3N and acts along the y axis...before you start drawing decide on your units...does 1N = 1cm, 1 inch, 1mm? (Not so ridiculous if the magnitudes of your forces are well into double figures)...once you pick your units draw the horizontal one (the 4N one) once this is drawn just draw the vertical one but as it's 3N make sure that the length of this line is 3/4 the length of the horizontal one (and points vertically)...you should end up with the two smaller sides of a right angled triangle.