Question

1. from the top of a 90.0m roof, a ball is thrown down with an initial velocity of 12.0 m/sec. find: (a) how long does the ball take to hit the ground? (b) what is the velocity of hitting the ground? (c) what time after throwing the ball down, the ball has a velocity of 25.0m/s. (d) what is the velocity of the ball 20m above the ground?

Explanation:

Part (a)

We use the kinematic equation for vertical motion: \( s = u t+\frac{1}{2} g t^{2} \), where \( s = 90.0\ m \) (displacement), \( u = 12.0\ m/s \) (initial velocity), \( g = 9.8\ m/s^{2} \) (acceleration due to gravity).

Step 1: Substitute values into the equation

\( 90 = 12t+\frac{1}{2}\times9.8\times t^{2} \)
\( 90 = 12t + 4.9t^{2} \)
\( 4.9t^{2}+12t - 90 = 0 \)

Step 2: Solve the quadratic equation \( ax^{2}+bx + c = 0 \) (here \( a = 4.9 \), \( b = 12 \), \( c=- 90 \))

Using the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \)
\( t=\frac{-12\pm\sqrt{12^{2}-4\times4.9\times(-90)}}{2\times4.9} \)
\( t=\frac{-12\pm\sqrt{144 + 1764}}{9.8} \)
\( t=\frac{-12\pm\sqrt{1908}}{9.8} \)
\( t=\frac{-12\pm43.68}{9.8} \)
We take the positive root (since time can't be negative)
\( t=\frac{-12 + 43.68}{9.8}=\frac{31.68}{9.8}\approx3.23\ s \)

Part (b)

We use the kinematic equation \( v = u+gt \)

Step 1: Substitute values

\( u = 12\ m/s \), \( g = 9.8\ m/s^{2} \), \( t\approx3.23\ s \)
\( v=12 + 9.8\times3.23 \)

Step 2: Calculate the value

\( v=12+31.654 = 43.654\approx43.7\ m/s \)

Part (c)

We use the kinematic equation \( v = u+gt \) and solve for \( t \)

Step 1: Rearrange the equation for \( t \)

\( t=\frac{v - u}{g} \)

Step 2: Substitute values \( v = 25\ m/s \), \( u = 12\ m/s \), \( g = 9.8\ m/s^{2} \)

\( t=\frac{25 - 12}{9.8}=\frac{13}{9.8}\approx1.33\ s \)

Part (d)

Answer:

s:
(a) \(\approx\boldsymbol{3.23\ s}\)

(b) \(\approx\boldsymbol{43.7\ m/s}\)

(c) \(\approx\boldsymbol{1.33\ s}\)

(d) \(\approx\boldsymbol{38.9\ m/s}\)