1. ## Projectiles again

A ball is kicked from a point A on the horizontal ground with an initial speed of 24.5 m/s at an angle a above the horizontal, where sina = 0.6. The ball hits the ground at point B.

a) Calculate the time of flight of the ball

b) Find the horizontal distance AB

c) Find the speed of the ball when it is at a height of 9.6 above the ground.

OK for a this is what I do but not sure if its right.

Vy = vsina - 9.8t
Vy = 24.5*0.6-0
Vy = 14.7

r = vsin(a)t - 4.9t²=0
r = 14.7*0.6t - 4.9t²=0

using the formula

(8.82+8.82)/9.8 = 1.8s

Could anybody confirm if this is done correctly?

for part b i get

AB = vsin(a)t - 4.9t²
=24.5*0.6*1.8-4.9(1.8)²
=26.46 - 15.876
= 10.584m

Don't know how to start part C :/

2. Ok you have a couple problems here Vy and Vx are the initial velocities

a,
you're right Vy = 14.7

so y(t) = 14.7 t -4.9t^2

the flight time is obtained by setting y = 0 and you get t = 3

b x(t) = Vx * t = 24.5cos(a) t = 24.5(.8) t = 19.6t

To get the horizontal distance compute x(3) = 58.8m

c. Set y = 9.6 and solve for t

use speed = [(dx/dt)^2 + (dy/dt)^2]^(1/2)

Note dx/dt = 19.6 for all t and dy/dt = 14.7 - 9.8t

3. Originally Posted by djmccabie
A ball is kicked from a point A on the horizontal ground with an initial speed of 24.5 m/s at an angle a above the horizontal, where sina = 0.6. The ball hits the ground at point B.

a) Calculate the time of flight of the ball

b) Find the horizontal distance AB

c) Find the speed of the ball when it is at a height of 9.6 above the ground.

OK for a this is what I do but not sure if its right.

Vy = vsina - 9.8t
Vy = 24.5*0.6-0
Vy = 14.7

r = vsin(a)t - 4.9t²=0
r = 14.7*0.6t - 4.9t²=0

using the formula

(8.82+8.82)/9.8 = 1.8s

Could anybody confirm if this is done correctly?

for part b i get

AB = vsin(a)t - 4.9t²
=24.5*0.6*1.8-4.9(1.8)²
=26.46 - 15.876
= 10.584m

Don't know how to start part C :/
a) $\displaystyle t = \frac{2v_0\sin{\theta}}{g}$

you should get $\displaystyle t = 3$ sec

b) $\displaystyle \Delta x = v_0\cos{\theta} \cdot t$

you should get $\displaystyle \Delta x = 58.8$ m

c) velocity in the x-direction remains constant ... $\displaystyle v_x = v_0 \cos{\theta}$

$\displaystyle v_y = \sqrt{(v_0\sin{\theta})^2 - 2g(\Delta y)}$

speed of the ball = $\displaystyle \sqrt{v_x^2 + v_y^2}$

you should get the speed to be $\displaystyle 20.3$ m/s

4. Hello, djmccabie!

A projectile problem requires two equations:

. . $\displaystyle x \;=\;(v_o\cos\theta)t$
. . $\displaystyle y \;=\;h_o + (v_o\sin\theta)t - 4.9t^2$

where: .$\displaystyle \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}$

A ball is kicked from a point $\displaystyle A$ on the horizontal ground
with an initial speed of 24.5 m/s at an angle $\displaystyle \theta$ above the horizontal,
where $\displaystyle \sin\theta = 0.6$. The ball hits the ground at point $\displaystyle B.$

We are given: .$\displaystyle h_o \:=\:0,\quad v_o \:=\:24.5,\quad \sin\theta \:=\:0.6,\;\cos\theta \:=\:0.8$

$\displaystyle \begin{array}{cccccc}\text{We have:} & x \:=\:(24.5)(0.8)t & \Rightarrow & x \:=\:19.6t & {\color{blue}(h)} \\ \text{And:} & y \:=\:(24.5)(0.6)t - 4.9t^2 & \Rightarrow & y \:=\:14.7t - 4.9t^2 & {\color{blue}(v)} \end{array}$

a) Calculate the time of flight of the ball
The "time of flight" is the period between the two instants
. . when the ball is on the ground $\displaystyle (y = 0).$ .We use (v).

We have: .$\displaystyle 14.7t - 4.9t^2 \:=\:0 \quad\Rightarrow\quad 4.9t(3-t) \:=\:0 \quad\Rightarrow\quad t \:=\:0,\,3$

The ball is on the ground at $\displaystyle t = 0$ (at kickoff).
The ball is once again on the ground at $\displaystyle t = 3.$

The time of flight is $\displaystyle \boxed{\text{3 seconds}}$

b) Find the horizontal distance $\displaystyle AB.$
For horizontal distance, we use (h).

When $\displaystyle t=3:\;\;x \:=\:(19.6)(3) \:=\:\boxed{58.8\text{ m}}$

c) Find the speed of the ball when it is at a height of 9.6 above the ground.
When is $\displaystyle y = 9.6\;?$

We have: .$\displaystyle 14.7t - 4.9t^2 \:=\:9.6 \quad\Rightarrow\quad 4.9t^2 - 14.7t + 9.6 \:=\:0$

Quadratic Formula: .$\displaystyle t \:=\:\frac{2.1 \pm\sqrt{0.57}}{1.4} \;\approx\;0.96,\:2.04$

We will use $\displaystyle t = 0.96$ .(Both give the same result.)

$\displaystyle \begin{array}{cccccc}\text{The horizontal speed is:} & s_h &=& x' &=& 19.6 \\ \text{The vertical speed is:} & s_v &=& y' &=& 14.7 - 9.8t \end{array}$

At $\displaystyle t = 0.96$, we have: .$\displaystyle s_h\:=\:19.6,\quad s_v \:=\:5.292$

Then: .$\displaystyle \text{speed} \:=\:\sqrt{(s_h)^2 + (s_v)^2} \:=\:\sqrt{(19.6)^2 + (5.292)^2} \;\approx\;\boxed{20.3\text{ m/sec}}$

Edit: .Ha! . . . Too slow again!
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