A projectile is fired with an initial speed of 37.0 m/s at an angle of 44.9 ∘ above the horizontal on a long flat firing range. Determine the maximum height reached by the projectile.
First of all, get the VERTICAL component of the initial velocity:
V = 37m/s (sin(44.9) = 26.117 m/s
When the Vertical velocity equals ZERO, the projectile has reached its highest point (apex)
since v= vo + at (vo = origianl velocity a = acceleration of gravity = -9.80 m /sec^2 t = time)
ZERO= 26.117 + (-9.80 t)
Solving for t
-26.117 = -9.8 t
26.117/9.8 = t = 2.665 seconds
NOW, use the position equation pf = po + vt +1/2at^2 pf=Final position (max height...apex)
po = 0
so:
pf = 26.117 (2.665) + 1/2 (-9.8)(2.665)^2 = 69.60 + (-3.48) = 66.1 m max height
+130518 The vertical displacement is given by:
h = v(sin theta)t + 1/2 (-9.8) t^2
And h' is the vertical velocity at any time.......this is 0 at the max ht.
And h' = vsin(theta) - 9/8t = 37sin(44.9) - 9.8t ...... set to 0
37*sin(44.9) = 9.8t
37*sin(44.9)/ 9.8 = t = 2.665 s
And putting this back into the position function, the max ht. =
h(2.665) = about 34.8 m
Here's the graph : https://www.desmos.com/calculator/smv5klqan8
First of all, get the VERTICAL component of the initial velocity:
V = 37m/s (sin(44.9) = 26.117 m/s
When the Vertical velocity equals ZERO, the projectile has reached its highest point (apex)
since v= vo + at (vo = origianl velocity a = acceleration of gravity = -9.80 m /sec^2 t = time)
ZERO= 26.117 + (-9.80 t)
Solving for t
-26.117 = -9.8 t
26.117/9.8 = t = 2.665 seconds
NOW, use the position equation pf = po + vt +1/2at^2 pf=Final position (max height...apex)
po = 0
so:
pf = 26.117 (2.665) + 1/2 (-9.8)(2.665)^2 = 69.60 + (-34.8) = 34.8 m max height
CORRECTED MATH ERROR IN LAST LINE
