Question

A bullet is fired from the ground at an angle 45 to the power of ring operator above the horizontal. What initial speed V subscript 0 must the bullet have in order to hit a point 125 meters high located 150 meters away? Use 10 space text m/s end text squared for gravity acceleration.

Answer 1

The initial speed of the bullet is 79.09 m/s

Step-by-step explanation:

Given data:

height of the building = 125 m

distance = 150 m

angle = 45°

The velocity of the bullet is:

`y=xtan\theta -(gx^(2) )/(2v^(2)cos\theta )`

Where x and y are horizontal and vertical distances.

`125=150*tan45-(9.8*150^(2) )/(2v^(2)*cos45 ) \\125=150-(220500)/(1.41v^(2) ) \\v=79.09m/s`

Answer 2

The initial speed of the bullet is 30 m/s

Step-by-step explanation:

Given;

angle of projection, θ = 45°

Apply kinematic equation, to determine time to reach vertical height of 125 meters;

Y = V
`_y`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