A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Fund the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?

Question

mrnood · Answer

First - resolve the initial velocity into horizontal and vertical components.
a) Horizontal motion is constant velocity - so s=ut
b) Vertical motion is acceleration under gravity- 
so s = ut + (gt^2)/2 so solve for s given t=3
c) use same equation as b - but solve for t (given s=10)

anonymous · Answer

what is s, u and t stands for?

mrnood · Answer

(NOTE - The equations above are independent in each axis - the distance s, and the value u are NOT the same in the 2 equations)

s = distance travelled 
u = initial velocity
t = time elapsed

anonymous · Answer

a) s=8m/s(3s)
s=24m/s^2
is it correct?

mrnood · Answer

NO - you must first resolve the velocity into horizontal and vertical components

anonymous · Answer

how?

mrnood · Answer

This is the basic concept of the work you are now doing - refer to your textbook. http://www.physicsclassroom.com/class/vectors/u3l1e.cfm The above link does the same for resolving forces - but it is IDENTICAL concept to resolving velocity. You will not be able to proceed in this topic till you understand the concept

mrnood · Answer

Note that the initial velocity is not horizontal - it is a 20 deg below horizontal.