Question

A stone is thrown horizontally outward from the top of a bridge, the stone is released 19.6meter above the street below.the initial velocity of the stone is 5.0m/s.determine the (a)total time that the stone is in the air
(b)magnitude and direction of the velocity"projectille just before the it strike the street.
(2)A projectile is fired with an initial speed of 113m/s at an angle of 60.0 degree above the horizontal from the top of a cliff 49.0m high.determine(a)the time to reach the maximum height
(b)maximum height above the base of the cliff reached by the projectile
(c) the total time it is in the air
(d)horizontal range of the projectile.plz help

Answer 1

If it is thrown horizontal, the horizontal speed u remains 5 m/s until it stops.
u = 5
v = - g t
h = 19.6 - .5 g t^2
or
0 = 19.6 - 4.9 t^2
t^2 = 19.6/4.9
t = 2 seconds in air

u = 5
v = -9.8 * 2 = -19.6 m/s

so speed at ground = sqrt (25 +19.6^2 )
tan angle below horizontal = 19.6/5

Answer 2

u = 11.3 cos 60 forever
Vi = 11.3 sin 60

v = 11.3 sin 60 - 9.8 t
v = 0 at top so
at top
t = 11.3 sin 60 /9.8
and also at top
h = 49 + Vi t - 4.9 t^2

now at h = 0
0 = 49 + Vi t - 4.9 t^2
solve quadratic for t when h = 0
then that is time in air
range = u t