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
Isn't it Velocity = x/t
To find the answers to the given questions, we'll need to break the problem down into smaller steps. Let's go through them one by one:
(a) The time of flight (t):
We can find the time of flight by using the equation for the y-displacement. The formula for the vertical displacement (y) of an object in projectile motion is:
y = v_0 * t * sin(θ) - (1/2) * g * t^2
In this equation, v_0 is the initial velocity, θ is the launch angle, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Since the stone is thrown downwards at an angle of 32.0° below the horizontal, the launch angle (θ) becomes -32.0°. Taking into account that the initial height (h) of the stone is 49.0 m, the vertical displacement (y) is -49.0 m.
Plugging in these values, we can solve the equation for t:
-49.0 m = v_0 * t * sin(-32.0°) - (1/2) * (9.8 m/s^2) * t^2
Simplifying and rearranging the equation, we get:
-4.9 t^2 + 15.81 t - 49 = 0
Now, we can solve this quadratic equation to find the values of t. Either you can use the quadratic formula or factorization to find the values of t.
(b) The initial speed (v_0):
To find the initial speed of the stone, we can use the equation for the horizontal displacement. The formula for the horizontal displacement (x) is:
x = v_0 * t * cos(θ)
Using the given values of x (69.3 m), t (which we found in part (a)), and θ (-32.0°), we can solve this equation for v_0.
69.3 m = v_0 * t * cos(-32.0°)
Simplifying the equation, we get:
v_0 = 69.3 m / (t * cos(-32.0°))
(c) Speed and angle of the velocity vector at impact:
To find the speed and angle of the velocity vector at impact, we can use the components of the velocity vector in the x and y directions. The magnitude of the velocity vector can be calculated using the Pythagorean theorem:
v = sqrt(v_x^2 + v_y^2)
The angle of the velocity vector with respect to the horizontal can be found using the arctangent function:
angle = arctan(v_y / v_x)
To calculate the components of the velocity vector, we can use the following equations:
v_x = v_0 * cos(θ)
v_y = v_0 * sin(θ) - g * t
Substituting the values of v_0, θ, t, and g into these equations, we can find the values of v_x and v_y. Then, we can calculate the magnitude of the velocity (v) and the angle (angle) using the formulas mentioned above.
I hope this explanation helps you understand how to solve the problem step by step. Let me know if you have any further questions!
you know the initial horizontal velocity was
Vh=Vicos32
distancehoizontal=vh/time
so time= vicos32/69.3
hf=hi-viSin32*vicos32/69.3 -1/2 g (vicos32/69.3)^2
or..0=69.3^2 - vi^2 sin32cos32-1/2 g vi^2 cos^2 32/69.3
you can calulate vi from that, then go back and calculate t.