A projectile is fired into the air from the top of a cliff of height h = 186 m above a valley. Its initial velocity is v0 = 64.8 m/s at an angle θ = 59° above the horizontal. Where does the projectile land? (Ignore any effects due to air resistance.)
vi vertical=64.8sin59
vi horizonal=64.8cos59
time in air:
hf=hi+vivertical*t- 1/2 g t^2
hf=0 hi=186 solve for time in air t.
then
horizonal distance=vihorizonatl*t
How would you find t? I can't get it.
Nevermind, I got it. Thank you!
To determine where the projectile lands, we can break down the problem into horizontal and vertical components.
First, let's find the time it takes for the projectile to reach the maximum height. We can use the vertical component of the initial velocity:
v0y = v0 * sin(θ)
= 64.8 m/s * sin(59°)
= 54.9999990635052 m/s
The time taken to reach the maximum height (t_max) can be found using the equation of motion:
t_max = v0y / g
where g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values:
t_max = 54.9999990635052 m/s / 9.8 m/s^2
= 5.6173469387755 s (approximately)
Next, let's find the total time of flight (t_total). Since the projectile goes up and then comes back down, the total flight time is twice the time to reach the maximum height:
t_total = 2 * t_max
= 2 * 5.6173469387755 s
= 11.234693877551 s (approximately)
Now, let's find the horizontal distance traveled during the total time of flight. We can use the horizontal component of the initial velocity:
v0x = v0 * cos(θ)
= 64.8 m/s * cos(59°)
= 33.7205864053955 m/s (approximately)
The horizontal distance (d) can be found using the equation:
d = v0x * t_total
= 33.7205864053955 m/s * 11.234693877551 s
= 378.729591836735 m (approximately)
Therefore, the projectile lands at a horizontal distance of approximately 378.73 meters from the base of the cliff.