A projectile is fired at an angle of 30ᵒ above the horizontal from the top of the cliff 600 ft high. The initial speed of the projectile is 2000 ft/s. how far will the projectile move horizontally before it hits the level ground at the base of the cliff?

Answer 1

A stone is dropped from a high altitude, and 3.00 s later another is projected vertically downward with a speed of 150 ft/ s. when and where will the second overtake the first?

Answer 2

let x and y be component by vertical motion only Vo (200ft/s) are;
Vxo=200Cos 30=173.2ft/s
Vyo=200sin 30=100ft/s
so we should find the time of flight (t) and this is determined by vertical motion only.
For vertical motion
h=Voyt+1/2gt(square)
h=Voyt+1/2gt(square
Here;
h=-600ft,Voy=100ft/s g=-9.8m/s=32.15ft/s t=?
-600=100t-1/2*32.15t(square)
solvng ths quadratc equation, t=3.75s(the negative value of T has no physical meaning since time intervals are always positive)
Distance in which projectile move horizontal;
R=Vox*t
R=173.2*3.75
R=649.5ft.