a trebuchet launches a projectile with an intial velocity of 15 m/s is launcged at an angle of 35 degrees. what will be the projectiles peak height
just do vertical problem
initial speed up = Vi = 15 sin 35 (around 8.5 m/s but you do)
v = Vi - g t = Vi - 9.81 t
when at the top,v = 0
so at the top
t = Vi / 9.81 (around .9 but you do)
then
h = Vi t - (1/2) g t^2
h = 8.5 *.9 - 4.9 (.9)^2 (my numbers are approximate)
Vo = 15m/s[35o].
Yo = 15*sin35 = 8.6 m/s. = Vert. component.
Y^2 = Yo^2 + 2g*h.
0 = 8.6^2 + (-19.6)h,
h = 3.77 m.
To determine the peak height of the projectile launched by the trebuchet, you can use the following formula:
H = (v² * sin²θ) / (2 * g)
Where:
H = peak height
v = initial velocity of the projectile (15 m/s)
θ = launch angle (35 degrees)
g = acceleration due to gravity (approximately 9.8 m/s²)
First, convert the launch angle from degrees to radians:
θ (in radians) = θ (in degrees) * π / 180
θ (in radians) = 35 * π / 180
Next, substitute the known values into the formula and calculate the peak height:
H = (15² * sin²(35° * π / 180)) / (2 * 9.8)
Using a calculator or computer software, calculate the value of this expression:
H ≈ 5.17 meters
Therefore, the projectile's peak height, when launched at an initial velocity of 15 m/s and an angle of 35 degrees, will be approximately 5.17 meters.