Posted by **HELP** on Monday, September 12, 2011 at 1:49am.

Consider a rock thrown off a bridge of height 78.5 m at an angle è = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 15.6 m/s. Find the following quantities:

(a) the maximum height reached by the rock

(b) the time it takes the rock to reach its maximum height

(c) the place where the rock lands

(d) the time at which the rock lands

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

magnitude

direction: Give your answer in degrees

- College Physics -
**Damon**, Monday, September 12, 2011 at 4:25am
u = horizontal speed = 15.6 cos 25 for the whole problem

Vi = initial speed up = 15.6 sin 25

in general

v = Vi = 9.8 t

h = 78 + Vi t - 4.9 t^2

for max height

v = 0 at top = Vi - 9.8 * time at top

height at top = 78 + Vi t - 4.9 t^2

- College Physics -
**Alex**, Sunday, June 30, 2013 at 7:13pm
A)2

b)5.7

C)87.7

D) 76.4

E)4.5

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