i have a few question that I have no idea how to answer?

Question

1. A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (6.5 i + 5.8 j) m/s, with i horizontal and j upward.
(a) To what maximum height does the ball rise?
m
(b) What total horizontal distance does the ball travel?
m
(c) What is the magnitude of the ball's velocity just before it hits the gound?
m/s
(d)What is the direction of the ball's velocity just before it hits the ground?
° relative to the horizontal

2.A particle leaves the origin with an initial velocity v = (1.00 i) m/s and a constant acceleration a = (-1.00 i - 0.200 j) m/s2
(a) What is the velocity of the particle when it reaches its maximum x coordinate?
m/s i + m/s j
(b) What is the position vector of the particle at this time?
m i + m j

Answer 1

1. let v = vx i + vy j

a. Consider y direction first

y = 1/2gt^2 + vy0 *t where g = -9.8 ms/^2 and vy0 = initial speed in y direction

since at y = 9.1 m, call this y1, we have vy = 5.8 m/s, we can write

y1 = (vy^2 - vy0^2)/(2*g)

ANd at the max height vy = 0 --> ymax =-vy0^2/(2g)

then y1 = vy^2/(2g)+ymax --> ymax = y1-vy^2/(2g)
ymax = 9.1 - (5.8)^2/(-2*9.8) = 10.82 m

b. Horizontal distance depends on time. We know max height and speed in x direction doesn't change since no forces act in x direction.

Now we know max height and that ymax = 1/2*g*t^2 +vy0*t
but we don't know vy0. We also now at vymax, the speed in y goes to zero so

0 = g*tmax + vy0 ---> vy0 = -g*tmax

Then ymax = 1/2*g*tmax^2 -g*tmax *tmax = 1/2*g*tmax

solving for tmax = 2.2 seconds. Total time in air is 2*tmax or 4.4 seconds.

Now ball travels a distance x = vt = 6.5* 4.4 m = 28.7 m

c. velocity at point of impact is v = (6.5 i - 5.8 j) m/s so magnitude is vmag = sqrt(6.5^2+(-5.8)^2) = 8.71 m/s

d. direction relative to horizontal angle = arctan(-5.8/6.5) = -41.7 deg below horizontal/

Can't give you answers for 2 - you seem to be missing info in the problem.

Answer 2

you did not follow your lessons...
:-)