A ball is fired from a height of 3.0 m above the ground,with a speed of 13 m/s and an angle of 22 degrees below the horizontal.(a) At what time does the ball hit the ground? (b) What is the impact velocity?
break the initial velocity into vertical and horiontal components.
Then use the vertical velocity in the distance equation..
3=Vvo*time - 4.9time^2 to solve for time.
Impact velocity will be the Vector sum of the initial vertical velocity and the final vertical velocity..
final velocityvertical=initialVv-g*t
Thanks
Mvi-mvf
To solve this problem, we need to break it down into two parts: (a) finding the time it takes for the ball to hit the ground and (b) finding the impact velocity.
(a) To find the time it takes for the ball to hit the ground, we can use the kinematic equation for vertical motion:
Δy = v₀y * t + (1/2) * g * t²
where Δy is the change in height (3.0 m in this case), v₀y is the vertical component of the initial velocity (v₀y = v₀ * sin(θ)), g is the acceleration due to gravity (-9.81 m/s²), and t is the time.
Plugging in the given values:
3.0 m = (13 m/s * sin(22°)) * t + (1/2) * (-9.81 m/s²) * t²
Simplifying the equation:
3.0 m = 4.67 m/s * t - 4.905 m/s² * t²
Rearranging and setting the equation equal to zero:
4.905 m/s² * t² - 4.67 m/s * t + 3.0 m = 0
We can now solve this quadratic equation by using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / (2a)
where a = 4.905 m/s², b = -4.67 m/s, and c = 3.0 m.
Using the quadratic formula, we get two possible solutions for t:
t₁ = 0.957 s
t₂ = 0.384 s
Since the ball hits the ground after it is launched, we can discard the negative solution. Therefore, the ball hits the ground at approximately 0.957 seconds.
(b) To find the impact velocity, we need to find the horizontal and vertical components of the velocity at the time of impact.
The horizontal component of velocity remains constant and is given by:
v₀x = v₀ * cos(θ)
Plugging in the given values:
v₀x = 13 m/s * cos(22°) ≈ 11.843 m/s
The vertical component of velocity changes due to the acceleration from gravity. We can use the equation:
v_y = v₀y + g * t
where v₀y is the vertical component of the initial velocity (v₀y = v₀ * sin(θ)), g is the acceleration due to gravity (-9.81 m/s²), and t is the time.
Plugging in the values:
v₀y = 13 m/s * sin(22°) ≈ -4.67 m/s
v_y = -4.67 m/s + (-9.81 m/s²) * 0.957 s ≈ -13.962 m/s
The impact velocity is the vector sum of the horizontal and vertical components:
V_impact = sqrt(v₀x² + v_y²)
Plugging in the values:
V_impact = sqrt((11.843 m/s)² + (-13.962 m/s)²) ≈ 18.321 m/s
Therefore, the impact velocity is approximately 18.321 m/s.