Question

A foot bell is kicked with a velocity of 20misec et en angle of 150 with the horizontal. The
maximum height reached by the bellis m. (g = 10m15​

Answer 1

Answer:

The question is ambiguous. A ball being thrown horizontally does not specify its direction.

While some understand the problem as that of a ball being thrown downward from a building, I took it as a ball being thrown upward from an inital height of 50 meters.

For this it is necessary to calculate the time taken for the ball to reach its maximum height, then use that time to calculate the maximum height, and finally combine the sum of the maximum height and the height of the building to calculate the time taken for the ball to reach the ground from its 50m + maximum height.

We know that when a ball is thrown upward, its velocity when it reaches its maximum height is zero.

So we have:

V(t) = u(t) - (0.5g)t^2

0 = 20 - 4.9 t^2

t = 2.02s, so the ball reaches its maximum height at 2.02 seconds

Now calculate the maximum height reached :

x(t) = u.t - (0.5g)t^2

x(2.02)= 20.(2.02) - 4.9(2.02)^2

x(2.02)= 20.4 m

So the ball reaches a maximum height of 20.4 meters.

The overall height of the ball is therefore 50 meters (height of building) + 20.4 meters, which is 70.4 meters.

Now we can work out the time taken for the ball to drop from 70.4 meters :

x(t) = (0.5g)t^2

70.4 = 4.9t^2

t= 3.8s

From the total height of 70. 4 meters, the ball takes 3.8 seconds to reach the ground. However if we take into account the time taken for the ball to have reached its maximum height when it was thrown, the total time from the moment the ball was thrown up until it hits the ground is 2.02s +3.8s

Which gives a total time of 5.82 seconds

Answer 2

The question is ambiguous. A ball being thrown horizontally does not specify its direction.

While some understand the problem as that of a ball being thrown downward from a building, I took it as a ball being thrown upward from an inital height of 50 meters.

For this it is necessary to calculate the time taken for the ball to reach its maximum height, then use that time to calculate the maximum height, and finally combine the sum of the maximum height and the height of the building to calculate the time taken for the ball to reach the ground from its 50m + maximum height.

We know that when a ball is thrown upward, its velocity when it reaches its maximum height is zero.

So we have:

V(t) = u(t) - (0.5g)t^2

0 = 20 - 4.9 t^2

t = 2.02s, so the ball reaches its maximum height at 2.02 seconds

Now calculate the maximum height reached :

x(t) = u.t - (0.5g)t^2

x(2.02)= 20.(2.02) - 4.9(2.02)^2

x(2.02)= 20.4 m

So the ball reaches a maximum height of 20.4 meters.

The overall height of the ball is therefore 50 meters (height of building) + 20.4 meters, which is 70.4 meters.

Now we can work out the time taken for the ball to drop from 70.4 meters :

x(t) = (0.5g)t^2

70.4 = 4.9t^2

t= 3.8s

From the total height of 70. 4 meters, the ball takes 3.8 seconds to reach the ground. However if we take into account the time taken for the ball to have reached its maximum height when it was thrown, the total time from the moment the ball was thrown up until it hits the ground is 2.02s +3.8s

Which gives a total time of 5.82 seconds