Fitzpatrick Projectile question (1 Viewer)

clintmyster · Jul 22, 2009

I don't have the book on me so if anyone could post up the question it would be great. Its question 22 of ex 25D, the one about the cricket ball. I tried to get the answer but just failed, its 7m/s. Here is my solution below. If someone could correct me on what I did wrong, that'd be great.

gurmies · Jul 22, 2009

A batsman hits a cricket ball 'off his toes' towards a fieldsman who is 65 m away. The ball reaches a maximum height of 4.9 m and the horizontal component of its velocity is 28 m/s. Find the constant speed with which the fieldsman must run forward, starting at the instant the ball is hit, in order to catch the ball at a height of 1.3 m above the ground. (Use g = 9.8)

Tanden · Jul 23, 2009

hey can someone here please help me with question 11. in 25d) fitz patrick.
the question is: A plane travelling at 630km/h releases a bomb from a height of 1000 metrs.
a) how long does it take to reach the ground?
b) what horizontal distance does the bomb travel? (1km/h = 5/18 m/s)

jet · Jul 23, 2009

Clintmyster, I'm not sure what you did, because I can't read it, but here is the way I did it:
$\text{Let the initial vertical component of velocity be } a \\ \therefore \ddot{y} = -g \\ \dot{y} = C - gt \\ t = 0, \dot{y} = a \\ \therefore \dot{y} = a - gt \\ y = at - \frac{gt^2}{2} + C \\ t = 0, y = 0 \\ y = at - \frac{gt^2}{2} \\ \text{Also, } \ddot{x} = 0 \\ \dot{x} = C \\ t = 0, \dot{x} = 28 \\ \therefore \dot{x} = 28t \\ x = 28t + C \\ t = 0, x = 0 \\ x = 28t \\ \text{Max height occurs when } \dot{y} = 0 \\ a - gt = 0 \\ t = \frac{a}{g} \\ y = \frac{a^2}{g} - \frac{ga^2}{2g^2} = 4.9 \\ a = \sqrt{9.8g} = 9.8 \\ \text{Thus, when } y = 1.3, at - \frac{gt^2}{2} = 1.3 \\ 9.8t - 4.9t^2 = 1.3 \\ 4.9t^2 - 9.8t + 1.3 = 0 \\ t = \frac{1}{7} \text{ or } t = \frac{13}{7} \\ \text{When } t = \frac{13}{7} \text{ the ball travels a horizontal distance of } 28 \cdot \frac{13}{7} = 52 \text{ m} \\ \text{The fieldsman must run a distance of } 65-52 = 13 \text{ m in } \frac{13}{7} \text{ seconds} \\ s = \frac{d}{t} \\ = \frac{13}{\frac{13}{7}} \\ = 7 s$

jet · Jul 23, 2009

Tanden:
$\text{Since the plane is travelling horizontally, there is no initial vertical velocity, i.e } \dot{x} = 175 \text{ ms}^{-1}, \dot{y}= 0 \\ \ddot{y} = -g \\ \dot{y} = C - gt \\ t = 0, \dot{y} = 0 \\ \dot{y} = -gt \\ y = C - \frac{gt^2}{2} \\ t = 0, y = 1000 \\ \therefore y = 1000 - \frac{gt^2}{2} \\ \ddot{x} = 0 \\ \dot{x} = C \\ t = 0, \dot{x} = 175 \\ \dot{x} = 175 \\ x = 175t + C \\ t = 0, x = 0 \\ x = 175t \\ \text{When } y = 0, 1000 - \frac{gt^2}{2} = 0 \\ gt^2 = 2000 \\ t^2 = 204.08 \\ t = 14.3 \text{ s} \\ \\ \text{When } t = 14.3, x = 175(14.3) \\ = 2500 \text{ m}$

Tanden · Jul 23, 2009

hey mate, thanks heaps for the help!

clintmyster · Jul 23, 2009

thanks mate, much appreciated

Fitzpatrick Projectile question (1 Viewer)

clintmyster

Prophet 9 FTW

gurmies

Drover

Tanden

New Member

jet

Banned

jet

Banned

Tanden

New Member

clintmyster

Prophet 9 FTW

Users Who Are Viewing This Thread (Users: 0, Guests: 1)