Consider a rock that is thrown off a bridge of height 58 m at an angle θ = 23° with respect to the horizontal

Consider a rock that is thrown off a bridge of height 58 m at an angle θ = 23° with respect to the horizontal. If the initial speed the rock is thrown is 13 m/s, find the following quantities.

(a) The time it takes the rock to reach its maximum height.

(b) The maximum height reached by the rock.

(c) The time at which the rock lands.

(d) The place where the rock lands.

(e) The velocity of the rock (magnitude and direction) just before it lands.
magnitude
direction

Horizontal component of velocity = (cos 23) x 13, = 11.967m/sec.
Vertical component = (sin 23) x 13, = 5.08m/sec.
a) Time to max. height = (v/g), = 5.08/9.8, = 0.5184 secs.
b) Height reached above ground = 58 + (v^2/2g), = 58 + 1.317, = 59.317 metres.
c) Time to landing = 0.5184 + sqrt. (2h/g), = 0.5184 + 3.44, = 3.959 secs.
d) (3.959 x 11.967) = 47.37m. from base of cliff.

Vertical velocity component at ground = sqrt. (2gh), = 34.1m/sec.

e) Velocity at ground = sqrt. (34.1^2 + 11.967^2), = 36.14m/sec.
Direction = atn. (11.967/34.1), = 19.33 deg. from vertical, or 70.67 from horizontal.