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
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
2 answers
A)2
b)5.7
C)87.7
D) 76.4
E)4.5
b)5.7
C)87.7
D) 76.4
E)4.5
1 answer