Question

Dave throws sweets upwards at a speed of 8 ms-1 from a first-floor window to his sister Carol who is standing at a second-floor window 3m directly above Dave. Carol misses catching the sweet as it passes her hand on the way up, but catches it on the way down.

a) Calculate the maximum height reached by the sweet above the point of projection from Dave's hand.

Answer 1

The maximum height reached by the sweet can be calculated using the equation h = (v^2)/(2g), where v is the initial vertical velocity and g is the acceleration due to gravity. Plugging in the given values, we find that the maximum height is approximately 3.27 meters.

Step-by-step explanation:

The maximum height reached by the sweet can be calculated using the equation:

h = (v^2)/(2g)

Where:

• h is the maximum height
• v is the initial vertical velocity
• g is the acceleration due to gravity

In this case, the initial vertical velocity is 8 m/s and the acceleration due to gravity is 9.8 m/s^2. Plugging in these values, we get:

h = (8^2)/(2*9.8) = 3.27 m

Therefore, the maximum height reached by the sweet is approximately 3.27 meters above the point of projection from Dave's hand.