fitzpatrick projectile questions :( (1 Viewer)

[norelle]

1. A particle projected from a point meets the horizontal plane through the point of projectin after travelling a horizontal distance a, and in the course of its trajectory attains a greatest height b above the point of projection.
Find the horizontal and vertical components of the velocity in terms of a and b. Show that when it has described a horizontal distance x, it has attained a height of 4bx(a-x) / a^2

2. A batsman hits a scricket ball "off his toes" towards a fieldsman who is 65m away. The ball reaches a maximum height of 4.9m and the horizontal component of its velocity is 28m/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 1.3m above the ground. (g=9.8)
ANS: 7m/s

3. Find the speed and direction of a particle which, when projected from a point 15m above the horizontal ground, just cleans the top of a wall 26.25 m high and 30m away.
ANS: 25m/s, 36*52*



Thanks so much ~~

 

[Timothy.Siu]

2.

x''=0
x'=28
x=28t

y''=-g
y'=-gt+Vsin@
y=-0.5gt²+Vtsin@

when y'=0 t=Vsin@/g
sub into y
4.9=v²sin²@/2g (they give 4.9 as max)
solving we get Vsin@=g

from x=28t
t=x/28
subbing into y. to get cartesian equation...
y=-0.5g(x²/28²)+Vsin@. (x/28)

we want to find when its 1.3m off the ground
1.3=-x²/160 +7x/20 (subbing in g=9.8 and Vsin@ from above)

x²-56x+208=0
solving..(x-52)(x-4)=0
x=52 and 4
therefore x=52m when the cricketer needs to catch it
subbing into x equation,
t=52/28
65-52=13m (how far fieldsman has to run)
13m in (52/28)seconds is 7m/s

next one..

this one means that ymax is 26.25 and x=30