Question

A cannon ball is launched at an angle of 48O with the ground. The cannon ball is ejected at a speed of 17.66 m/s. It hits a target that is 30 meters away at a height of 1.75 m. What is the time for the ball to reach the target?
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Answer 1

Answer:

A cannon ball is launched at an angle of 48° with the ground. The cannon ball is ejected at a speed of 17.66 m/s. It hits a target that is 30 meters away at a height of 1.75 m. What is the time for the ball to reach the target?

Explanation:

I hope you understand

Answer 2

The time required for the ball to reach the target will be 2.68 seconds.

Given:

Angle = 48°

Initial Velocity = 17.66 m/s

Height = 1.75 m

Range = 30 m

To Find:

The time required for the ball to reach the target.

Solution:

It's fairly simple to find the answer to this question, as seen below.

We have an equation connecting the time and range of a projectile.

$gT^{2}$ = 2Rtanθ

$T^{2}$ = $\frac{2Rtan}{g}$

T = $\sqrt{\frac{2Rtan}{g} }$

On substituting the given values,

T = $\sqrt{\frac{2*30*1.2}{10} }$

We know that tan 48 = 1.2

Therefore,

T = $\sqrt{7.2}$

T = 2.68 seconds.

Hence, the time required for the ball to reach target is 2.68 seconds.

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