sph3u1 - unit 1 kinematics
10. which statement is true for a soccer kick at an angle of 15°?
a. the ball has no initial velocity in the y - direction
b. the acceleration changes direction after the ball reaches the highest point
c. the ball will accelerate in the x - direction after the ball is kicked.
d. there is a positive displacement in the y - direction when the ball lands on ground
e. none of above
short/long response - show all your work including the formula applied, remember to include units and directions where applicable. check your significant figures in your answers.
11. a cannonball is fired horizontally with an initial velocity of 65 m/s off a 25 m high cliff. assume there is no air resistance, how far from the cliff will the cannon ball land? include a diagram. 5 marks - t
12. convert the following position - time graph into a velocity time graph 4 marks - c

Question

Accepted Answer

### 10.
# Explanation:
## Step1: Analyze option a
A soccer - kick at an angle has an initial y - velocity component $v_{0y}=v_0\sin	heta$. Since $	heta = 15^{\circ}$, $v_{0y}
eq0$, so a is false.
## Step2: Analyze option b
The acceleration during projectile motion is due to gravity $g = 9.8\ m/s^{2}$ (downward) and it does not change direction during the motion. So b is false.
## Step3: Analyze option c
In the absence of air - resistance, there is no acceleration in the x - direction ($a_x = 0$) for projectile motion. So c is false.
## Step4: Analyze option d
When the ball lands on the ground, the displacement in the y - direction is zero (assuming the starting and ending y - positions are the same). So d is false.
# Answer:
e. None of above

### 11.
# Explanation:
## Step1: Analyze the vertical motion
The vertical displacement of the cannon - ball is given by $y = y_0+v_{0y}t-\frac{1}{2}gt^{2}$. Here, $y_0 = 25\ m$, $v_{0y}=0\ m/s$, and $y = 0\ m$. So, $0 = 25+0	imes t-\frac{1}{2}	imes9.8t^{2}$.
## Step2: Solve for time t
Rearranging the equation $4.9t^{2}=25$, we get $t=\sqrt{\frac{25}{4.9}}\approx2.26\ s$.
## Step3: Analyze the horizontal motion
In the horizontal direction, $x = v_{0x}t$. Given $v_{0x}=65\ m/s$ and $t = 2.26\ s$, then $x=65	imes2.26 = 146.9\ m$.
# Answer:
The cannon - ball will land approximately $147\ m$ from the cliff.

### 12.
The velocity is the slope of the position - time graph.
- When the position - time graph is decreasing, the slope is negative, so the velocity is negative.
- When the position - time graph is increasing, the slope is positive, so the velocity is positive.
- When the position - time graph has a horizontal section (slope = 0), the velocity is 0.
We can divide the time interval into sections:
1. In the first section, the position - time graph is decreasing with a non - zero slope, so the velocity is negative and constant (since the slope is constant).
2. When the position - time graph crosses the x - axis, the slope is 0, so the velocity is 0.
3. In the second section, the position - time graph is decreasing again with a non - zero slope, so the velocity is negative and constant.
4. In the last section, the position - time graph is increasing, so the velocity is positive and constant.
We can sketch the velocity - time graph accordingly. The velocity - time graph will have horizontal lines (constant velocity) in different sections corresponding to the constant - slope sections of the position - time graph, and a value of 0 at the point where the slope of the position - time graph is 0.

It is not possible to provide a numerical graph here, but the general procedure is to calculate the slopes of different linear sections of the position - time graph to get the values of velocity for the velocity - time graph.

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QUESTION IMAGE

Question

Explanation:

10.

Step1: Analyze option a

Step2: Analyze option b

Step3: Analyze option c

Step4: Analyze option d

Step1: Analyze the vertical motion

Step2: Solve for time t

Step3: Analyze the horizontal motion

Answer:

11.