A stone is aimed at a cliff of height h with an initial speed of 62 m/s directed 45° above the horizontal, as shown in the Figure below. The stone strikes at A, 6.56 s after launching. What is the height of the cliff?
Can someone please tell me which equation(s) I need to use for this?
Not knowing where A is, it is going to be general.
Break the initial velocity into vertical and horizontal components. Then, using those, and the time of flight, you have two equations which should solve the problem:
Horizontal
distance= vih*time
Vertical:
height= viv*time - 4.9 time^2
To solve this problem, we can use the equations of projectile motion. Specifically, we will use the equations for the horizontal and vertical components of motion separately.
For the horizontal motion, the equation we will use is:
Δx = V0x * t
where Δx is the horizontal displacement, V0x is the initial horizontal velocity, and t is the time of flight (which is 6.56 s in this case).
For the vertical motion, we will use the equation:
Δy = V0y * t + (1/2) * g * t^2
where Δy is the vertical displacement, V0y is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the stone is launched at an angle of 45° above the horizontal, we can determine the initial horizontal and vertical velocities using trigonometry:
V0x = V0 * cos(45°)
V0y = V0 * sin(45°)
where V0 is the initial speed (62 m/s in this case).
To find the height of the cliff (h), we need to find the vertical displacement (Δy) when the stone strikes at point A. Therefore, we set Δy equal to h:
h = V0y * t + (1/2) * g * t^2
Now we can substitute the values we know into the equation and solve for h.