can someone please help me qith this question i am having trouble please

Question

Peter is throwing a ball at an initial velocity of 28 feet per second and an angle of 56 with the horizontal from a height of 4 feet.

a)how far away will the ball land?

b)what will be its max height?

someone please help

Answer 1

We take coordinate systemxoy (origin o on the ground). Then equations of motion: x = 28cos56°t (ft),

v_x = 2cos56° (ft/sec); y = 4 + 28sin56°t - 16t² (ft) and v_y = 28sin56° - 32t (ft/sec).

To get time of motion we set y = 0, 4 + 28sin56°t - 16t² = 0, we get t = 1.606 sec.

Then a) Far away distance is equal x-coordinate of point of landing, s = 28cos56°·1.606 = 25.15 ft

b)At highest point v_y = 0 or 28sin56° - 32t = 0, t = 28sin56°/32 = .7254 sec. Then maximum heght is y-coordinate at this moment: h_max= 4 + 28sin56°· 0.7254 - 16·0.7254² = 12.42 ft