From a height of 25 meters a ball is thrown vertically upwards at a velocity of 5 meters per second.

a.What time in seconds will the ball reach its maximum height?

b.What is its maximum height?

c.When will the ball strike the ground?

d.What will be the ball’s velocity when it strikes the ground?

can someone help me please!

v = 5-9.8t

h = 25+5t-4.9t^2

Now look for when v=0 or h=0

Of course, I can help you with these questions. Let's break them down and find the answers step by step.

a. To find the time when the ball reaches its maximum height, we can use the formula for vertical motion:

\(v_f = v_i + at\)

In this case, the initial velocity (vi) is 5 m/s, and the final velocity (vf) at the maximum height is 0 m/s (since the ball momentarily stops at the top). The acceleration (a) due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of the initial velocity). Let's plug the values into the equation:

\(0 = 5 - 9.8t\)

Rearranging the equation, we have:

\(9.8t = 5\)

\(t = \frac{5}{9.8} \approx 0.51\) seconds

So, the ball will reach its maximum height after approximately 0.51 seconds.

b. The maximum height can be found using the equation:

\(h = v_i t + \frac{1}{2} at^2\)

In this case, the initial velocity (vi) is 5 m/s, and the time (t) is the value we found in part a. The acceleration (a) due to gravity is -9.8 m/s^2. Plugging the values into the equation:

\(h = 5 \cdot 0.51 + \frac{1}{2} \cdot (-9.8) \cdot (0.51)^2\)

Simplifying the equation, we get:

\(h \approx 1.28\) meters

So, the maximum height reached by the ball is approximately 1.28 meters.

c. To determine when the ball will strike the ground, we can use another equation for vertical motion:

\(h = v_i t + \frac{1}{2} a t^2\)

This time, the initial height (h) is 25 meters, the initial velocity (vi) is 5 m/s, and the acceleration (a) is -9.8 m/s^2. We want to find the time (t) when the height becomes 0 (as the ball reaches the ground). The equation becomes:

\(0 = 5t + \frac{1}{2} \cdot (-9.8) \cdot t^2\)

Rearranging the equation, we have:

\(4.9t^2 - 5t = 0\)

Now we can factor out t:

\(t(4.9t - 5) = 0\)

This means that either \(t = 0\) (which is the starting point) or \(4.9t - 5 = 0\). Solving the equation \(4.9t - 5 = 0\), we find:

\(4.9t = 5\)

\(t = \frac{5}{4.9} \approx 1.02\) seconds

Thus, the ball will strike the ground after approximately 1.02 seconds.

d. To find the velocity when the ball strikes the ground, we can use:

\(v_f = v_i + at\)

In this case, the initial velocity (vi) is 5 m/s, the time (t) is the value found in part c, and the acceleration (a) is -9.8 m/s^2. Plugging the values into the equation:

\(v_f = 5 + (-9.8) \cdot 1.02\)

Simplifying the equation, we get:

\(v_f \approx -4.96\) m/s

The negative sign indicates that the velocity is in the opposite direction of the initial velocity. So, the ball's velocity when it strikes the ground is approximately -4.96 m/s.