A ball is thrown upward from the top of a 30 m high building. The ball takes \( 2,4 \mathrm{~s} \) to reach its highest point. Ignore the effects of air resistance. Calculate: a) the velocity with which the ball left the thrower's hand. b) the total time taken for the ball to reach the ground.

To solve the problem, we'll analyze the motion of the ball using kinematic equations under the influence of gravity. ### Part a) **Calculating the Initial Velocity** **Given:** - Time to reach the highest point, \( t = 2.4 \, \text{s} \) - Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) (acting downward) **Approach:** At the highest point, the final velocity (\( v \)) of the ball is \( 0 \, \text{m/s} \). We can use the equation: \[ v = u + a t \] Where: - \( v \) is the final velocity - \( u \) is the initial velocity - \( a \) is the acceleration (here, \( a = -g = -9.8 \, \text{m/s}^2 \) because gravity acts downward) - \( t \) is the time **Calculation:** \[ 0 = u - g t \] \[ u = g t \] \[ u = 9.8 \, \text{m/s}^2 \times 2.4 \, \text{s} = 23.52 \, \text{m/s} \] **Answer:** The ball was thrown upward with an initial velocity of **23.52  m/s**. ### Part b) **Calculating the Total Time to Reach the Ground** **Given:** - Initial height of the building, \( h = 30 \, \text{m} \) - Initial velocity upward, \( u = 23.52 \, \text{m/s} \) - Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \) **Approach:** We'll use the kinematic equation for vertical motion: \[ s = u t + \frac{1}{2} a t^2 \] Here, \( s = -30 \, \text{m} \) (since the ball is falling 30 meters below the initial point), \( a = -9.8 \, \text{m/s}^2 \), and \( u = 23.52 \, \text{m/s} \). Substituting the values: \[ -30 = 23.52 t - \frac{1}{2} \times 9.8 t^2 \] \[ -30 = 23.52 t - 4.9 t^2 \] Rearranging: \[ 4.9 t^2 - 23.52 t - 30 = 0 \] Solving this quadratic equation for \( t \): \[ t = \frac{23.52 \pm \sqrt{(23.52)^2 + 4 \times 4.9 \times 30}}{2 \times 4.9} \] \[ t = \frac{23.52 \pm \sqrt{553.1904 + 588}}{9.8} \] \[ t = \frac{23.52 \pm \sqrt{1141.1904}}{9.8} \] \[ t = \frac{23.52 \pm 33.78}{9.8} \] Taking the positive root (since time cannot be negative): \[ t = \frac{23.52 + 33.78}{9.8} \approx \frac{57.3}{9.8} \approx 5.85 \, \text{s} \] **Answer:** The total time taken for the ball to reach the ground is approximately **5.85  seconds**. ### Summary - **a)** Initial velocity: **23.52 m/s upward** - **b)** Total time to reach the ground: **5.85 seconds**