projectiles questionX2 (1 Viewer)

[nazi jew]

this is the exact wording of the questions:

1) A particle projected from a point meets the horizontal plane through the point of projection 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 attainted a height of {[4BX(a-X)] / a^2}.... (g=10)

2)A batsmen hits a cricket 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 foward, 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)

solutions: 1: (a/2).sqrt(g/2b).sqrt(2gb)
2: 7m/s

the questions are fine logically and diagramatically... i just have trouble with the algebra... (P.S. our class isnt learning any formulae for this topic, bar the '6 equations of motion'... would i need formulae (i.e. maximum height) in order to do these questions?).... thanks.

 

[haboozin]

jeez do they want any more words in that question


oooppsss

i get

100bX(a - x)/a^2

everything else is good except for the 100...

btw Q is an angle

basicly i found that a = v^2sinQ.CosQ/5
and b = v^2sin^2Q/20

and played around with the formulae's

i also made v^2 = 20b/sin^2Q

 

[..:''ooo]

for the second question, find how long it takes for the projectile to be 1.3m above ground (which is the same time the runner must run). and find the horizontal distance at this time.
then use speed = distance / time. (where distance is 65 - the range found b4).

hint : formulate your max height with given informations, and everything should simplify out.

 

[nazi jew]

thanks boys