A projectile is shot at a 37 degree angle from the edge of a cliff h = 305 m above ground level with an initial speed of v0 = 125 m/s
how much time will it take to hit the ground
vertical : velocityinitial Viv=125sin37
hfinal=ho+Viv*t-4.9t^2
ho=0,hf=-125, Viv above, solve for t.
To find the time it takes for the projectile to hit the ground, we can use the kinematic equations of motion.
First, let's break down the initial velocity (v0) into its horizontal (v₀x) and vertical (v₀y) components. Since the projectile is shot at an angle of 37 degrees above the horizontal, we can use trigonometry to calculate these components.
The horizontal component (v₀x) can be determined using cosine:
v₀x = v₀ * cosθ
The vertical component (v₀y) can be determined using sine:
v₀y = v₀ * sinθ
where:
v₀ is the initial velocity (125 m/s in this case)
θ is the launch angle (37 degrees in this case)
Given that the initial velocity v₀ = 125 m/s and the launch angle θ = 37 degrees, we can calculate:
v₀x = 125 * cos(37)
v₀y = 125 * sin(37)
Now that we have the horizontal and vertical components, we can find the time it takes for the projectile to hit the ground by considering the vertical motion.
The formula to calculate the time of flight (t) in the vertical direction is given as:
t = (2 * v₀y) / g
where:
g is the acceleration due to gravity (approximately 9.8 m/s²)
Using the equation above, we substitute the calculated value of v₀y to find the time t it takes for the projectile to hit the ground.