Example Question - initial velocity
Here are examples of questions we've helped users solve.
Determining the Time and Maximum Height of a Projectiles Launch
<p>To determine the time at which the projectile reaches its maximum height, we need to find when the velocity is zero. The velocity is the derivative of the position function \( y(t) \).</p> <p>The position function given in the question is:</p> <p>\[ y(t) = -16t^2 + 64t + 80 \]</p> <p>To find the velocity, we differentiate \( y(t) \) with respect to \( t \):</p> <p>\[ v(t) = y'(t) = \frac{d}{dt}(-16t^2 + 64t + 80) = -32t + 64 \]</p> <p>We set the velocity to zero and solve for \( t \):</p> <p>\[ 0 = -32t + 64 \]</p> <p>\[ 32t = 64 \]</p> <p>\[ t = 2 \]</p> <p>The projectile reaches its maximum height at \( t = 2 \) seconds.</p> <p>Now we calculate the maximum height by substituting \( t = 2 \) back into the position function \( y(t) \):</p> <p>\[ y(2) = -16(2)^2 + 64(2) + 80 \]</p> <p>\[ y(2) = -16(4) + 128 + 80 \]</p> <p>\[ y(2) = -64 + 128 + 80 \]</p> <p>\[ y(2) = 144 \]</p> <p>Therefore, the maximum height reached by the projectile is 144 feet.</p>
Calculating the Time for a Ball to Reach Zero Speed when Thrown Upward
<p>To calculate the time for the ball to reach zero speed, we can use the following kinematic equation for uniformly accelerated motion:</p> \[ v_f = v_i + a \cdot t \] <p>Where:</p> <p>\( v_f \) is the final velocity (0 m/s when the ball reaches its maximum height and stops momentarily)</p> <p>\( v_i \) is the initial velocity (4.5 m/s)</p> <p>\( a \) is the acceleration (acceleration due to gravity; for a ball thrown upward this is -9.8 m/s\(^2\), since it acts downward)</p> <p>\( t \) is the time in seconds</p> <p>We can set \( v_f \) to 0 and solve for \( t \):</p> \[ 0 = 4.5 \text{ m/s} - (9.8 \text{ m/s}^2) \cdot t \] \[ 9.8 \text{ m/s}^2 \cdot t = 4.5 \text{ m/s} \] \[ t = \frac{4.5 \text{ m/s}}{9.8 \text{ m/s}^2} \] \[ t \approx 0.459 \text{ s} \] <p>So, the ball will take approximately 0.459 seconds to reach zero speed.</p>
Calculating Distance Traveled by an Accelerating Car from Rest
<p>The distance traveled by an object under constant acceleration can be found using the equation:</p> <p>\[ s = ut + \frac{1}{2}at^2 \]</p> <p>where:</p> <p>\[ u \] is the initial velocity (in this case, 0 m/s, since the car starts from rest),</p> <p>\[ a \] is the constant acceleration (3 m/s^2), and</p> <p>\[ t \] is the time (4 seconds).</p> <p>Plugging in the values, we get:</p> <p>\[ s = (0) \cdot 4 + \frac{1}{2} \cdot 3 \cdot (4)^2 \]</p> <p>\[ s = 0 + \frac{1}{2} \cdot 3 \cdot 16 \]</p> <p>\[ s = 0 + \frac{1}{2} \cdot 48 \]</p> <p>\[ s = 24 \text{ meters} \]</p> <p>Thus, the distance traveled by the car is 24 meters.</p>
Calculating Final Velocity of a Bicycle Under Constant Acceleration
<p>v_{final} = v_{initial} + a * t</p> <p>v_{final} = 5 m/s + (2 m/s^2 * 3 s)</p> <p>v_{final} = 5 m/s + 6 m/s</p> <p>v_{final} = 11 m/s</p>
Calculating Final Velocity of a Car After Uniform Acceleration
<p>$v = u + at$</p> <p>Where:</p> <p>$v$ is the final velocity,</p> <p>$u$ is the initial velocity (which is $0$ m/s since the car starts from rest),</p> <p>$a$ is the acceleration ($2 \ m/s^2$),</p> <p>$t$ is the time ($5$ seconds).</p> <p>Substitute the given values into the equation:</p> <p>$v = 0 + (2 \ m/s^2)(5 \ s)$</p> <p>$v = 10 \ m/s$</p> <p>Therefore, the car's velocity after $5$ seconds is $10 \ m/s$.</p>
Calculating Launch Angle for Projectile Motion
The image contains a text in Italian that talks about a projectile being fired with an initial velocity, and it asks to calculate the launch angle such that the projectile can hit a target placed a certain distance away. Unfortunately, the full details of the problem are not completely visible, but I can provide a generic approach for such a problem. To solve for the launch angle (\(\theta\)) of a projectile given the initial velocity (\(v_0\)) and the distance to the target (\(R\)), we can use the following formula, assuming no air resistance and that the launch and target points are at the same vertical level: \[ R = \frac{v_0^2}{g} \sin(2 \theta) \] Where \( R \) is the range or distance to the target, \( v_0 \) is the initial velocity, \( g \) is the acceleration due to gravity (which is approximately \( 9.81 m/s^2 \) on Earth), and \( \theta \) is the launch angle. Rearrange this equation to solve for \( \theta \): \[ \sin(2 \theta) = \frac{R g}{v_0^2} \] Take the inverse sine to find \( 2\theta \): \[ 2\theta = \sin^{-1}\left(\frac{R g}{v_0^2}\right) \] Then, divide by 2 to find the angle \( \theta \): \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{R g}{v_0^2}\right) \] Now, input the given values (\( v_0 = 130 m/s \) and \( R = 1100 m \)), and solve for \( \theta \): \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1100 \times 9.81}{(130)^2}\right) \] \[ \theta = \frac{1}{2} \sin^{-1}\left(\frac{10791}{16900}\right) \] \[ \theta = \frac{1}{2} \sin^{-1}(0.6383) \] \[ \theta \approx \frac{1}{2} \times 39.86^\circ \] \[ \theta \approx 19.93^\circ \] The launch angle needed to hit the target 1100 meters away with an initial velocity of 130 m/s is approximately \( 19.93^\circ \). Keep in mind that due to the nature of the sin function there are two possible solutions for \( \theta \) in the range [0°, 90°], as sin(θ) = sin(180° - θ). Therefore, the other possible angle would be \( 90^\circ - 19.93^\circ = 70.07^\circ \). These two angles represent the low and high trajectories that both reach the target.
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