Question

A ball is projected upwards from a bridge and hits the ground 5 seconds after it was originally projected its initial velocity is 15 m/s and it’s final velocity is 34 m/s calculate the height of the bridge.Max height = 11.48 m

Answer 1

Given:

• Time, t =5 seconds

,

• Initial velocity, u = 15 m/s

,

• Final velocity, v = 34 m/s

Given that the ball is projected upwards from a bridge, let's find the height of the bridge.

Let's first find the height from the top of the bridge to the maximum height.

To find the height apply the formula:

`v^2=u^2-2as`

Where:

v is the final velocity = 0 m/s (velocity at max height).

u is the initial velocity = 15 m/s

a is acceleration due to gravity = 9.8 m/s²

s is the height.

`\begin{gathered} 0^2=15^2-2(9.8)s \\ \\ 0=225-19.6s \\ \\ 19.6s=225 \\ \\ s=(225)/(19.6) \\ \\ s=11.48\text{ m} \end{gathered}`

The distance from the bridge to the maximum height is 11.48 m.

Now, let's find the maximum height to the ground.

`v^2=u^2+2as`

Where:

v is the final velocity = 34 m/s

u is the initial velocity = 0 m/s (velocity at the top).

a is acceleration due to gravity = 9.8 m/s²

s is the maximum height.

Thus, we have:

`\begin{gathered} 34^2=0^2+2(9.8)s \\ \\ 1156=19.6s \\ \\ s=(1156)/(19.6) \\ \\ s=58.9\approx59\text{ m} \end{gathered}`

The maximum height is 59 meters.

To find the height of the bridge, we have:

Height of bridge = Maximum height - Distance from top of bridge to max height.

Height of bridge = 59m - 11.48m = 47.52 m

Therefore, the height of the bridge is 47.52 meters.

• ANSWER:

47.52 meters.