Answer:
d = 20.25 m
Explanation:
h = 6m, V = 14 m/s, θ=14°
⇒ Vx=V × Cosθ=13.58 m/s and Vy = V × Sinθ=3.39 m/s
to find time t by h=-Vy t + 1/2 g t² (after putting value and simplifying) (-ve for downward motion)
4.9 t² - 3.39 t - 6 =0
⇒ t=1.5 sec (ignoring -ve root)
Now Vx = d/t
13.5 m/s = d / 1.5 s ⇒ d = 20.25 m
Answer:
The horizontal projectile range is 20.44m
Explanation:
To determine the range of the projectile we need to first determine the total time of flight.
Using the vertical component of the motion.
The first phase of flight (moving up)
Using equation of motion.
h = ut + 0.5at^2 + h0 .....1
a = -g (acceleration due to gravity acting against motion) = -9.8m/s^2
u = vertical initial speed = 14sin14°
h0 = initial height = 6m
h = 0 (on the x axis y = 0)
Equation 1, becomes;
h = ut - 0.5gt^2 + h0
Substituting the values
0 = 14sin14(t) - 4.9t^2 + 6
3.39t - 4.9t^2 +6 = 0
Solving the quadratic equation, we have;
t = 1.505s or t = -0.814s
time cannot be negative, so
t = 1.505s
Since the total time of flight is 1.505s
The range R = Vx × t
Vx = horizontal speed = 14cos14 m/s
R = 14cos14 × 1.505
Range R = 20.44 m
