Daniel B. answered 10/10/22
A retired computer professional to teach math, physics
Let
g = 9.81 m/s² be gravitation acceleration,
v(t) be the (vector) velocity of the rock after some time t.
So v(0) = v0.
Let t1 be the time it takes the rock to hit the ground.
The assignment is to compute t1 and v(t1).
v0 can be decomposed into a vector sum of
its horizontal component of magnitude v0cos(θ0) and
its vertical component of magnitude v0sin(θ0).
While the horizontal component remains constant, the vertical component
is constantly reduced by gravity.
So after some time t the vertical component of v(t) is
v0sin(θ0) - gt (1)
The horizontal component of position (with respect to the starting position) is
v0cos(θ0)t
because horizontally the rock moves at constant velocity.
The vertical position of the rock (with respect to the starting position) is
v0sin(θ0)t - gt²/2
(If you do not know where this comes from let me know.)
We are given that at the sought time t1 the vertical position is -20, so
v0sin(θ0)t1 - gt1²/2 = -20
Substituting actual numbers we get the quadratic equation
4.9t1² - 17.4t1 - 20 = 0
t1 = (17.4 ± √(17.4² + 196.2))/9.81
This gives us two solutions
t1 = -0.5, and t1 = 4
The solution t1 = -0.5 does not correspond to our physical constraints,
while t1 = 4s is the answer to question (a)
We can plug that into (1) to get the vertical component of the velocity v(t1) as
29×sin(37°) - 9.81×4 = -21.8 m/s
Note that it is negative, because it is oriented downward.
The horizontal component of v(t1) is the constant
v0cos(θ0) = 29×cos(37°) = 23.2 m
The tangent of θ is the ration between the two components
tan(θ) = -21.8/23.2 = -0.94
Therefore θ = -43.2°
(The angle is negative because it points below the horizontal.)
The magnitude of the velocity (i.e., speed) is the Euclidean sum of the two components
|v(t1)| = √(21.8² + 23.2²) = 31.8 m/s