promotion image of download ymail app
Promoted
Mcc asked in Science & MathematicsPhysics · 2 months ago

Projectile Question?

A particle P is projected vertically upwards with speed 25 m s−1

from a point 3 m above horizontal

ground.

i) Find the time taken for P to reach its greatest height. i found the answer ,it is 2.5 s

ii)Find the length of time for which P is higher than 23 m above the ground. (how do i do this ?)

iii) P is higher than h m above the ground for 1 second. Find h.

i dont understand how to tackle these questions. :D

3 Answers

Relevance
  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    ii) there are a couple of routes. I would set up the quadratic describing vertical motion

    h = h0 + v*t - ½gt²

    and plug in the knowns:

    23 m = 3m + 25m/s*t - ½*9.8m/s²*t²

    4.9t² - 25t + 20 = 0

    quadratic with roots at

    t = 0.993 s

    and t = 4.11 s

    The ball is above 23 m BETWEEN these two times, or for about 3.1 seconds. ◄

    You could instead find the maximum height H and then solve

    H - 23m = ½gt²

    for t and then double it (up then down). I'll leave that for you if you're interested.

    iii) Now I think we WILL need to find the maximum height.

    H = h0 + v² / 2g = 3m + (25m/s)² / 19.6m/s² = 34.9 m

    If the ball was above height h for 1 second, then it was rising from h for ½ second and falling back to h for ½ second.

    How far does it fall from H in ½ second?

    y = ½gt² = ½ * 9.8m/s² * (0.5s)² = 1.225 m

    so

    h = 34.9m - 1.23 m = 33.7 m ◄

    Hope this helps!

    • Commenter avatarLogin to reply the answers
  • Anonymous
    2 months ago

    i) find the time t taken for P to reach its greatest height

    time t = Voy/g = 25/9.806 = 2.549 sec

    ii) find the time t' for which P is higher than 23 m above the ground

    up time tu :

    (23-ho) = 25*tu-4.903t^2

    tu = (25-√

    25^2-4.903*4*(23-3) ) / 9.806 = 0.994 sec

    Δtu = t-tu = 2.549-0.994 = 1.556 sec

    down time td :

    Hmax = ho+Voy^2/2g = 3+25^2/19.612 = 34.87 m

    td = √2*(Hmax-23)/g = √2*(34.87-23)/9.806 = 1.556 sec

    t' = Δtu+td = 1.556+1.556 = 3.112 sec

    or more simply : t' = 2*Δtu = 1.556*2 = 3.112 sec

    iii) P is higher than h'' m above the ground for t'' = 1 sec : find h''.

    thanks to symmetry of the trajectory near the apex :

    h'' = Hmax-g*(t''/2)^2/2 = 34.87-9.806*0.5^2/2 = 33.64 m

    • Commenter avatarLogin to reply the answers
  • 2 months ago

    Ii) time taken = sq.root ( 2h/g ) = sq.root(2x23/10) = 2.144 sec

    Iii) height h = 25x1+1/2x10x1= = 30m.

    • Commenter avatarLogin to reply the answers
Still have questions? Get your answers by asking now.