The range of the projectile is maximum at **angle 45°**, the range is **10.2m**, the time to reach the maximum height is **0.72 s** and the maximum height is **2.6 m**

<h3>

**What is a Projectile ?**</h3>

The **stone, or object or anything** projected in a trajectory path is known as a projectile.

In the given two-dimensional projectile motion problem involving a ball starting and ending at the same height. the projectile leaves the ground with an initial velocity of 10 m/s at some angle Ф with the horizontal,

**(a)** The value of the angle Ф (in degrees) which will makes the range of the projectile maximum is **angle 45° ** because Range = u²sin2Ф ÷ g

At Ф = 45

Range = u²sin90° ÷ g

where sin 90 = 1

So, Range = u² / g which is the maximum range.

**(b)** The range at this angle will be

Range = 10² ÷ 9.8

Range = 100 / 9.8

**Range = 10.2 m**

**(c)** To calculate how long it takes the ball to reach maximum height, we will use the formula

t = usinФ ÷ g

t = 10 sin 45 ÷ 9.8

**t = 0.72s**

**(d)** The maximum height will be

H = u²sin²Ф ÷ 2g

H = 10²(sin45)² ÷ 2 × 9.8

H = 100 × 0.5 ÷ 19.6

H = 50 ÷ 19.6

**H = 2.55 m**

Therefore, the range of the projectile is maximum at **angle 45°**, the range is **10.2m**, the time to reach the maximum height is **0.72 s** and the maximum height is **2.6 m**

Learn more about **Projectile** here: brainly.com/question/12870645

#SPJ1