I was wondering if you could help me solve this projectile motion problem.
A projectile is launched with an initial speed of 14.5 m/s at an angle of 35° above the horizontal. The object lands at the same height from which it was launched. Air resistance is negligible. Determine
a) Maximum height: 3.5m
b) The projectiles horizontal displacement when it hits the ground
c) How long it will take to reach the maximum height. 0.85s
I've done everything except b. The problem is that you're only given the time and velocity after. I tried d=vt and that gave me a wrong answer. Am I leaving something out?
Problem is the textbook gives me the formula: ΔDx=vi^2/g * Sin2(θ), which works however we did not learn the material that way.
you can see how that formula was achieved if you read the article on trajectory in wikipedia.
Maybe your textbook also explains how the formula was derived. In any case, just use it to get your answer!
To solve part b) of the problem, you need to find the horizontal displacement of the projectile when it hits the ground. Since the initial height and the final height are the same, the vertical displacement is zero. Therefore, you can use the formula for horizontal displacement:
ΔDx = Vx * t
where ΔDx is the horizontal displacement, Vx is the horizontal component of the initial velocity, and t is the time of flight.
To find Vx, you can use the formula:
Vx = Vi * cos(θ)
where Vi is the initial velocity and θ is the launch angle.
To find t, you can use the equation for the time of flight of a projectile:
t = 2 * Vy / g
where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity (approximately 9.8 m/s^2).
First, calculate Vx using the given values:
Vx = 14.5 m/s * cos(35°) ≈ 14.5 m/s * 0.819 ≈ 11.89 m/s
Next, calculate t:
t = 2 * Vy / g
To find Vy, you can use the formula:
Vy = Vi * sin(θ)
Vy = 14.5 m/s * sin(35°) ≈ 14.5 m/s * 0.574 ≈ 8.32 m/s
t = 2 * 8.32 m/s / 9.8 m/s^2 ≈ 1.69 s
Finally, substitute Vx and t into the formula for horizontal displacement:
ΔDx = (11.89 m/s) * (1.69 s) ≈ 20.08 m
Therefore, the horizontal displacement of the projectile when it hits the ground is approximately 20.08 m.