posted by anne .
Suppose the stone is thrown at an angle of 32.0° below the horizontal from the same building (h = 49.0 m) as in the example above. If it strikes the ground 69.3 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
(a) the time of flight s
(b) the initial speed m/s
(c) the speed and angle of the velocity vector with respect to the horizontal at impact
angle ° below the horizontal
you know the initial horizontal velocity was
so time= vicos32/69.3
hf=hi-viSin32*vicos32/69.3 -1/2 g (vicos32/69.3)^2
or..0=69.3^2 - vi^2 sin32cos32-1/2 g vi^2 cos^2 32/69.3
you can calulate vi from that, then go back and calculate t.
can u explain it in another way
Isn't it Velocity = x/t