Suppose the stone is thrown at an angle of 32.0° below the horizontal from the same building (h = 49.0 m) as in the example above. If it strikes the ground 69.3 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)
(a) the time of flight s
(b) the initial speed m/s
(c) the speed and angle of the velocity vector with respect to the horizontal at impact
speed m/s
angle ° below the horizontal
heres two answers i got for the same question.
physics - bobpursley, Friday, May 20, 2016 at 4:49pm
break the initial velocity into two componnets;
horizontal: Vo*cos40
vertical: -Vo*sin40 - 9.8 t
a. hf=-42=-vo*sin40*t-1/2 9.8 t^2
d=36.7=vo*cos40t
solve for vo*t in the second in terms of t. Put that in the first equaiton.
b. knowing t, use the second equation to solve for vo.
physics - Damon, Friday, May 20, 2016 at 4:55pm
42 = Vo t + 4.9 t^2
Vo = speed sin 40
36.7 = u t
u = speed cos 40
so
speed * t = 36.7/cos 40
so
speed * sin 40 * t = 36.7 tan 40
Vo t = 36.7 tan 40
so back to
42 = Vo t + 4.9 t^2
now
42 = 36.7 tan 40 + 4.9 t^2
solve for t, time in the air, and go back and get the rest
To solve this problem, we can break it down into several steps. Let's start with part (a):
(a) The time of flight (t) can be found using the equation for the vertical displacement (y). Since the stone is thrown at an angle below the horizontal, we need to consider both the horizontal and vertical components of its motion.
The equation for the horizontal displacement (x) is given by: x = v0 * t * cos(theta)
The equation for vertical displacement (y) is given by: y = v0 * t * sin(theta) - (1/2) * g * t^2
In the example above, we found the value of x to be 69.3 m. Since the stone is thrown from the same building, the height (h) is still 49.0 m.
Using the equation for x, we can rewrite it as: t = x / (v0 * cos(theta))
Substituting this value of t into the equation for y, we get: h = (v0 * sin(theta) * (x / (v0 * cos(theta)))) - (1/2) * g * (x / (v0 * cos(theta)))^2
Now we can solve for t.
(b) To find the initial speed (v0), we need to consider the equation for the horizontal displacement (x) again: x = v0 * t * cos(theta)
We already know the values of x and t. Solve this equation for v0 to find the initial speed.
(c) To find the speed and angle of the velocity vector at impact, we need to find the horizontal and vertical components of the final velocity.
Horizontal component of velocity (Vx) can be calculated using: Vx = v0 * cos(theta)
Vertical component of velocity (Vy) can be calculated using: Vy = v0 * sin(theta) - g * t
The speed (V) can be calculated using the Pythagorean theorem: V = sqrt(Vx^2 + Vy^2)
The angle (alpha) below the horizontal can be calculated using: alpha = tan^(-1)(Vy / Vx)
These calculations will help you find the required values in parts (a), (b), and (c) of the problem.