Question

I need help on a physic hw

A ball is shot at 14.5m/s at an angle of 50.0˚ above the horizontal, which is north. (One picture is fine.) (Include direction in a, b, and c!)
a) What is its initial vertical velocity?
b) What is its vertical velocity at the top of its path?
c) What its horizontal velocity at the top of its path?
d) How long will it take it to travel 15.0m North?
e) How high will it be in part d)?
f) If it is shot from ground level, when will it land on the ground?

Just letter B and C only

Answer 1

The vertical velocity at the top of its path is zero, which is why that is the top of the path.

Ignoring wind resistance, the horizontal velocity at the top of the path is the same it started with, 14.5 cos (50)

Answer 2

okay so all I do is answer the question and not doing any solving??

Answer 3

c will require a calculation: 14.5 cos (50)

Answer 4

I need help on letter E

Answer 5

height will be (initial vertical velocity)(time) -(1/2) (grav accel) (time)^2.

h = 14.5 sin (50) t - (1/2)(9.8)(t)^2

find t from when it has reached zero vertical velocity

0 = 14.5 sin (50) - (9.8)(t)

There you are....