A ball is thrown horizontally from the roof of building 50.m tall and lands 45m from the base. What was the balls initial speed
1. A ball is thrown horizontally from the roof of a building 50. m tall and lands
45 m from the base. What was the ball’s initial speed?
7.92 m/s
Well, it seems the ball definitely had some air time, but unfortunately, it seems to have misplaced its speedometer. The ball's initial speed remains a mystery!
To find the initial speed of the ball, we can use the kinematic equation for horizontal motion, which is:
distance = velocity × time
In this case, the ball was thrown horizontally, so there is no vertical acceleration. Therefore, the time of flight for the ball will be the same as the time it takes for the ball to fall vertically from the roof to the ground.
We can calculate the time using the equation for vertical motion:
distance = (1/2) × acceleration × time^2
Since the ball falls vertically, the distance is equal to the height of the building: distance = 50 m.
The acceleration due to gravity can be considered as -9.8 m/s^2, taking into account that downward direction is negative.
Plugging in the values, we have:
50 = (1/2) × (-9.8) × time^2
Simplifying the equation:
100 = -9.8 × time^2
Dividing both sides by -9.8:
-10.204 = time^2
Taking the square root of both sides:
time ≈ √(-10.204) (Note: We only consider the positive square root since time cannot be negative)
time ≈ 3.2 seconds (rounded to one decimal place)
Now that we have the time of flight, we can calculate the initial speed (velocity) using the horizontal distance traveled by the ball:
distance = velocity × time
Plugging in the values:
45 = velocity × 3.2
Dividing both sides by 3.2:
velocity ≈ 14.06 m/s (rounded to two decimal places)
Therefore, the ball's initial speed is approximately 14.06 m/s.
time of flight ... 50 = 1/2 g t^2
initial speed = 45 m / t