A canon is fired at an angle `theta(0le theta le(pi)/(2))` with the horizontal. If 'v' is the intial velocity of the canon ball, the height 'h' of the ball at time 't', ignoring wind resistance, is given by `h=(v sin theta)t-4.9t^(2)`.
(a) Will be ball return to the ground?
(b) How far will the ball have travelled horizontally at the time it hits the ground, assuming there are no forces in the horizontal direction?
(c ) Determine `'theta'` so that the horizontal range of the ball is maximum.

Answer

To solve the problem step by step, we will address each part of the question systematically.

### Given:
The height \( h \) of the cannonball at time \( t \) is given by the equation:
\[ h = (v \sin \theta)t - 4.9t^2 \]

### (a) Will the ball return to the ground?

To determine if the ball returns to the ground, we need to find when the height \( h \) becomes zero:
\[ 0 = (v \sin \theta)t - 4.9t^2 \]

Factoring out \( t \):
\[ t \left( v \sin \theta - 4.9t \right) = 0 \]

This gives us two solutions:
1. \( t = 0 \) (the time when the ball is fired)
2. \( v \sin \theta - 4.9t = 0 \)

From the second equation, we can solve for \( t \):
\[ 4.9t = v \sin \theta \implies t = \frac{v \sin \theta}{4.9} \]

Since \( t \) is positive, the ball will return to the ground after some time. Thus, **yes**, the ball will return to the ground.

### (b) How far will the ball have traveled horizontally at the time it hits the ground?

The horizontal distance \( R \) traveled by the ball can be calculated using the horizontal velocity and the total time of flight. The horizontal component of the velocity is:
\[ v_x = v \cos \theta \]

The total time of flight is \( t = \frac{v \sin \theta}{4.9} \). Therefore, the horizontal distance \( R \) is given by:
\[ R = v_x \cdot t = (v \cos \theta) \left( \frac{v \sin \theta}{4.9} \right) \]

Substituting the values:
\[ R = \frac{v^2 \sin \theta \cos \theta}{4.9} \]

Using the identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \):
\[ R = \frac{v^2}{9.8} \sin(2\theta) \]

Thus, the horizontal distance traveled when the ball hits the ground is:
\[ R = \frac{v^2 \sin(2\theta)}{9.8} \]

### (c) Determine \( \theta \) so that the horizontal range of the ball is maximum.

To maximize the range \( R \), we need to differentiate \( R \) with respect to \( \theta \):
\[ R = \frac{v^2}{9.8} \sin(2\theta) \]

Differentiating \( R \):
\[ \frac{dR}{d\theta} = \frac{v^2}{9.8} \cdot 2 \cos(2\theta) \]

Setting the derivative equal to zero for maximization:
\[ 2 \cos(2\theta) = 0 \implies \cos(2\theta) = 0 \]

This occurs when:
\[ 2\theta = 90^\circ + n \cdot 180^\circ \quad (n \in \mathbb{Z}) \]

Thus:
\[ \theta = 45^\circ + n \cdot 90^\circ \]

Since \( \theta \) must be in the range \( [0, \frac{\pi}{2}] \), the only valid solution is:
\[ \theta = 45^\circ \]

### Summary of Answers:
(a) Yes, the ball will return to the ground.
(b) The horizontal distance traveled when the ball hits the ground is \( R = \frac{v^2 \sin(2\theta)}{9.8} \).
(c) The angle \( \theta \) for maximum range is \( 45^\circ \).