Question

a ball is thrown upward from ground level with an initial velocity of 26.4 m/s. a ball first reach a height of 4.05 meters. at what time did the ball first reach a height of 4.05 meters? at what time did the ball have a velocity of 9.75 m/s. at what time was it thrown (above where it was thrown meters)?

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v = v_0+at$ and $v^{2}-v_{0}^{2}=2ah$. Here, $a=-g=- 9.8\ m/s^{2}$, $v_0 = 26.4\ m/s$.

Step2: First, find the time when the ball reaches the maximum - height

At the maximum - height, $v = 0$. Using $v = v_0+at$, we can solve for $t$.
$t=\frac{v - v_0}{a}=\frac{0 - 26.4}{-9.8}\ s\approx2.69\ s$.

Step3: Then, find the height at the maximum - height

Using $v^{2}-v_{0}^{2}=2ah$, when $v = 0$, we have $h=\frac{v^{2}-v_{0}^{2}}{2a}=\frac{0-(26.4)^{2}}{2\times(-9.8)}\ m\approx35.5\ m$.

Step4: Now, find the time when the velocity is $9.75\ m/s$

Using $v = v_0+at$, we solve for $t$. $t=\frac{v - v_0}{a}$. When $v = 9.75\ m/s$, $t=\frac{9.75 - 26.4}{-9.8}\ s\approx1.7\ s$.

Step5: Find the height when the velocity is $9.75\ m/s$

Using $h = v_0t+\frac{1}{2}at^{2}$, substituting $v_0 = 26.4\ m/s$, $a=-9.8\ m/s^{2}$, and $t = 1.7\ s$.
$h=26.4\times1.7+\frac{1}{2}\times(-9.8)\times(1.7)^{2}$
$h = 44.88-14.161=30.719\ m$.

Answer:

The time when the velocity is $9.75\ m/s$ is approximately $1.7\ s$ and the height at that time is approximately $30.7\ m$.