Question

A ball is thrown with an initial velocity of 10ms-1 from a cliff 45m high. What is the angle of the ball with the horizontal just before it hits the ground?

Answer 1

what does this mean: 10ms-1 ?

Answer 2

the equation becomes:

h(t) = -4.9 t^2 + Vi t + 45

The derivative of it should indicate the angle.
s(t) = -9.8 t + Vi

or something along that nature.

Answer 3

10 metres per second

Answer 4

do we know the initial angle of the velocity? since this equation is an up and down equation more than a forward

Answer 5

h(t) = -4.9 t^2 +10 sin(a) t + 45 express the amount of time in the air.

v(t) = 10 cos(a) t = how far out it travels