Answer:
The initial speed of the bullet is 30 m/s
Explanation:
Given;
angle of projection, θ = 45°
Apply kinematic equation, to determine time to reach vertical height of 125 meters;
Y = V[tex]_y[/tex]t + ¹/₂gt²
At maximum height, v = 0
125 = ¹/₂ x 10t²
125 = 5t²
t² = 125/5
t² = 25
t = √25
t = 5 seconds
Finally, the magnitude of the initial speed of the bullet to reach 150 meters horizontal distance
X = V₀t + ¹/₂gt²
gravity has little or no influence in horizontal displacement
X = V₀t
150 = V₀ x 5
V₀ = 150/5
V₀ = 30 m/s
Therefore, the initial speed of the bullet is 30 m/s
Answer:
The initial speed of the bullet is 79.09 m/s
Explanation:
Given data:
height of the building = 125 m
distance = 150 m
angle = 45°
The velocity of the bullet is:
[tex]y=xtan\theta -\frac{gx^{2} }{2v^{2}cos\theta }[/tex]
Where x and y are horizontal and vertical distances.
[tex]125=150*tan45-\frac{9.8*150^{2} }{2v^{2}*cos45 } \\125=150-\frac{220500}{1.41v^{2} } \\v=79.09m/s[/tex]
