Cliff That Is 157 M Height. Kick A Rock Into River At A Velocity Of 17.0 M/s. How Far From The

A ball thrown horizontally from the top of a building 55m high strikes the ground at a point 35m from the building. What is the (a) time to reach the ground, (b) the initial speed of the ball, and (c) the velocity by which the ball will strike the ground?

For the first bit you can use $s=\frac{1}{2}gt^2$so $55=\frac{1}{2}*9.8*t^2$$t^2=11.22\,\,\,\, so\,\,\,\,\, t=3.35s$The key to these type of questions is finding t first like this because it is the same for the vertical and horizontal components.For part (B) you can use $s=ut+\frac{1}{2}at^2$Since acceleration a is zero in the horizontal component this becomes $s=ut$We know s (given) and we know t (from part A) so you can go on and calculate u. This is the horizontal component of velocity $v_x.$For the last part we consider first the vertical component of motion so we can use $v^2=u^2+2as$$u=0$$a=9.8\,m.s^{-2}$$s=55m$You can get $v_y$Now we know both horizontal and vertical components we can find the resultant using Pythagoras:$v_{res}^{2}=v_{x}^{2}+v_{y}^{2}$So a tip when handling ballistic type questions like these is to find t first as this is the common factor between the horizontal and vertical components of the particle's trajectory, then use the other equations of motion.