Question

0 - Incomplete Projectile Motion: Look again at chapter 3, sections 7 and 8. Problem 1: Problem 3.29 of Giancoli lan bah Problem 2: I throw a ball, mass m-5.0kg, at an initial speed of 3.0m/s, at an initial height of 1.7m, but at an angle of 37' with respect to the horizontal. It lands on the ground, at a final height of zero. (a) What is the initial velocity? (This is a VECTOR. Define ti as the horizontal, and tj as the vertical.) (b) What is the time when the ball obtains its maximum height? (c) What is the maximum height of the ball? (d) What is the velocity at the maximum height? It is NOT zero!E 6 (e) What is the acceleration at the maximum height? (D) When it reaches the maximum height, where is the ball? (g) What is the velocity JUST BEFORE it hits the ground? (h) What is the speed JUST BEFORE it hits the ground? (i) Where does it land? ts maximum height? 1.7m ax 9.8'? 1.7 1.27s+ 2.14 0.8 ee

Answer 1

(a)

v_o = initial speed of launch = 3 m/s

= angle of initial velocity of launch with horizontal = 37

v_o = initial speed of launch = 3 m/s theta = angle of initial velocity of launch with horizontal = 37 v_ox = component of initial velocity along x-direction = v_o Cos theta = 3 cos 37 = 2.4 m/s v_oy = component of initial velocity along y-direction = v_0 sin theta = 3 sin 37 = 1.8 m/s In unit vector notation, the velocity is given as v vector = (v_ox) i^cap + (v_oy)j^cap v vector = 2.4 i^cap + 1.8 j^cap consider the motion along the vertical direction or Y-direction v_oy = component of initial velocity along y-direction = 1.8 m/s v_fy = final velocity at the highest point = 0 m/s a_y = acceleration = -9.8 m/s^2 t = time taken to reach the maximum height using the equation v_fy = v_oy + at 0 = 1.8 + (- 9.8) t t = 0.18 sec Y_o = initial height at the time of launch = 1.7 m

v_ox = component of initial velocity along x-direction = v_o Cos = 3 Cos37 = 2.4 m/s

v_o = initial speed of launch = 3 m/s theta = angle of initial velocity of launch with horizontal = 37 v_ox = component of initial velocity along x-direction = v_o Cos theta = 3 cos 37 = 2.4 m/s v_oy = component of initial velocity along y-direction = v_0 sin theta = 3 sin 37 = 1.8 m/s In unit vector notation, the velocity is given as v vector = (v_ox) i^cap + (v_oy)j^cap v vector = 2.4 i^cap + 1.8 j^cap consider the motion along the vertical direction or Y-direction v_oy = component of initial velocity along y-direction = 1.8 m/s v_fy = final velocity at the highest point = 0 m/s a_y = acceleration = -9.8 m/s^2 t = time taken to reach the maximum height using the equation v_fy = v_oy + at 0 = 1.8 + (- 9.8) t t = 0.18 sec Y_o = initial height at the time of launch = 1.7 m

v_oy = component of initial velocity along y-direction = v_o Sin = 3 Sin37 = 1.8 m/s

v_o = initial speed of launch = 3 m/s theta = angle of initial velocity of launch with horizontal = 37 v_ox = component of initial velocity along x-direction = v_o Cos theta = 3 cos 37 = 2.4 m/s v_oy = component of initial velocity along y-direction = v_0 sin theta = 3 sin 37 = 1.8 m/s In unit vector notation, the velocity is given as v vector = (v_ox) i^cap + (v_oy)j^cap v vector = 2.4 i^cap + 1.8 j^cap consider the motion along the vertical direction or Y-direction v_oy = component of initial velocity along y-direction = 1.8 m/s v_fy = final velocity at the highest point = 0 m/s a_y = acceleration = -9.8 m/s^2 t = time taken to reach the maximum height using the equation v_fy = v_oy + at 0 = 1.8 + (- 9.8) t t = 0.18 sec Y_o = initial height at the time of launch = 1.7 m

In unit vector notation , the velocity is given as

$\underset{v}{\rightarrow}$ = (v_ox ) $\hat{i}$ + (v_oy ) $\hat{j}$

$\underset{v}{\rightarrow}$ = 2.4 $\hat{i}$ + 1.8 $\hat{j}$

b)

consider the motion along the vertical direction or Y-direction

v_oy = component of initial velocity along y-direction = 1.8 m/s

v_fy = final velocity at the highest point = 0 m/s

a_y = acceleration = - 9.8 m/s²

t = time taken to reach the maximum height

using the equation

v_fy = v_oy + a t

0 = 1.8 + (- 9.8) t

t = 0.18 sec

c)

Y_o = initial height at the time of launch = 1.7 m

Y = height at the highest point = ?

using the equation

v_fy² = v_oy² + 2 a (Y - Y_o)

0² = 1.8² + 2 (- 9.8) (Y - 1.7)

Y = 1.87 m

d)

at the maximum height , we have

v_fy = component of velocity along y-direction = 0 m/s

v_fx = component of velocity along x-direction = v_ox = 2.4 m/s

so velocity at highest point is given as

$\underset{v_{f}}{\rightarrow}$ = (v_fx ) $\hat{i}$ + (v_fy ) $\hat{j}$

$\underset{v_{f}}{\rightarrow}$ = (0) $\hat{i}$ + (2.4) $\hat{j}$

e)

at the maximum height :

a_y = acceleration along the Y-direction = - 9.8 m/s²

a_x = acceleration along the x-direction = 0 m/s²

acceleration is given as

$\underset{a}{\rightarrow}$ = a_x $\hat{i}$ + a_y $\hat{j}$

$\underset{a}{\rightarrow}$ = 0 $\hat{i}$ - 9.8 $\hat{j}$

f)

the ball is at maximum height at t = 0.184 sec

position of the ball along x-direction is given as

x = x_o + v_ox t + (0.5) a_x t²

x = 0 + 2.4 (0.184) + (0.5) (0) (0.184)²

x = 0.44 m

y = 1.87 m

hence the position is given as

$\underset{r}{\rightarrow}$ = x $\hat{i}$ + y $\hat{i}$

$\underset{r}{\rightarrow}$ = 0.44 $\hat{i}$ + 1.87 $\hat{i}$

g)

consider the motion along the y-direction from top to bottom

Y_o = initial position = 1.7 m

Y = final position on ground = 0 m

using the equation

Y = Y_o + v_oy t + (0.5) a_y t²

0 = 1.7 + 1.8 t + (0.5) (- 9.8) t²

t = 0.8 sec

$\underset{v_{o}}{\rightarrow}$ = initial velocity = 2.4 $\hat{i}$ + 1.8 $\hat{j}$

$\underset{v_{f}}{\rightarrow}$ = final velocity = ?

$\underset{a}{\rightarrow}$ = acceleration = 0 $\hat{i}$ - 9.8 $\hat{j}$

t = 0.8 sec

using the equation

$\underset{v_{f}}{\rightarrow}$ = $\underset{v_{o}}{\rightarrow}$ + $\underset{a}{\rightarrow}$ t

$\underset{v_{f}}{\rightarrow}$ = (2.4 $\hat{i}$ + 1.8 $\hat{j}$ ) + ( 0 $\hat{i}$ - 9.8 $\hat{j}$ ) (0.8)

$\underset{v_{f}}{\rightarrow}$ = 2.4 $\hat{i}$ - 6.04 $\hat{j}$

h)

speed is given as

| $\underset{v_{f}}{\rightarrow}$ |= sqrt((2.4)² + (- 6.04)²)

| $\underset{v_{f}}{\rightarrow}$ |= 6.5 m/s

i)

final position is given as

$\underset{r_{f}}{\rightarrow}$ = $\underset{r_{o}}{\rightarrow}$ + $\underset{v_{o}}{\rightarrow}$ t + (0.5) $\underset{a}{\rightarrow}$ t²

$\underset{r_{f}}{\rightarrow}$ = (0 $\hat{i}$ + 1.7 $\hat{j}$ ) + (2.4 $\hat{i}$ + 1.8 $\hat{j}$ ) (0.8) + (0.5) (0 $\hat{i}$ - 9.8 $\hat{j}$ ) (0.8)²

$\underset{r_{f}}{\rightarrow}$ = (0 $\hat{i}$ + 1.7 $\hat{j}$ ) + 1.92 $\hat{i}$ + 1.44 $\hat{j}$ + (0 $\hat{i}$ - 3.14 $\hat{j}$ )

$\underset{r_{f}}{\rightarrow}$ = 1.92 $\hat{i}$ + 0 j