Question

Two seconds after being projected from ground level, a projectile is displace 40mhorizontally and 53mvertically above its launching point. What are the
(a)horizontal and
(b)vertical components of the initial velocity of the projectile?
(c)At the instant when the projectile achieves its maximum height above ground level, how far is it displace horizontally from the launch point.

Answer 1

horizontal displacement is constant, vertical is dependant on r(t)=vt-5t^2 (s=s'+
at 2 secs we get
r(2)=2v-5*4=53 v=(53+20)/2 v=36.5m/s this is the vertical component
horizontally we get that r(t)=vt
r(2)=2v=40 v=20m/s

to solve the last you could either apply x(axis) or the derivative

Answer 2

d/dt r(t) = v-10t
36.5-10t=0
t=3.65

Answer 3

horizontal its 26.5m/s

Answer 4

sorry i mean vertical lol

Answer 5

dunno man.. i still think it is 36.5
r(2)=v*2-5*4=53
53=2*v-20
2v=73
v=36.5

Answer 6

this is more of a physics question

Answer 7

oh, to the one with the question, anything u did not understand feel free to ask dude!

Answer 8

isn't it 53m/2s
26.5/s

Answer 9

forgot to add the deacceleration! gravity influencing over the vertical speed

Answer 10

so v=v0+at
0=26.5-9.8t
t=2.7s

Answer 11

its asking to find the velocity which is the displacement

Answer 12

u using the constant motion formula, but the projectile is submitted to gravity

Answer 13

so acceleration of gravity only affects y when your looking for the y value?

Answer 14

so when do you use v=v0+at and when do you use x-xo=v0t+1/2at^2?

Answer 15

vertical: r(t)=r0+v0*t-1/2*g*t^2
horizontal: r(t)=r0+v0*t