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post the questions?
yea man, ill do some for you if you post them. what unit is it on?
horizontal component: vx = dx/t , where dx is the change in distance in the x direction, vx is the x component of the velocity of the projectile
vertical component: dy = vy*t - .5gt^2 , where dy is the change in distance in the y direction, vy is the y component of the velocity, and g is the acceleration of gravity
in both cases, t = time the projectile spends in the air and it is the same in both the x and y directions, as the time taken for the projectile to travel along the ground is intuitively going to be the same amount of time it spends in the air.
vy = vsin(angle) and vx = vcos(angle) , provided the angle was given between the horizontal plane and the trajectory of the projectile.
of course, i'm guessing that these physics questions have something to do with projectile motion, and it seems to be the number one thing most people have trouble with. from what i've given, cross out what you know, circle what you need to find, and rearrange equations to solve for the shit you've circled.
first, you need to find out how long he's in the air for.
d = vt - .5gt^2 where d = 0 because he starts on the ground, soars through the air, and lands on the ground , v = 60sin5
0 = 60sin75 - .5(9.81)(t)^2
t = sqrt ( (60sin75) / 4.9 )
t = 3.44
second, you need to find the maximum vertical distance during his flight.
intuitively, since he's shot from the ground, heads up, peaks, and dips back down (which takes a total of 3.44 seconds) we now know that he's at his peak at the 1.72s mark (the midpoint of his flight). so we can say that it takes him 1.72 seconds to reach his peak height.
using the same equation as in question 1, where this time we're solving for d:-
d = vt - .5gt^2
d = 60sin75 - .5(9.81)(1.72)^2
d = 43.5m