a, b, and c are fine
d ...
hi there, i need apointer on this question, i've completed a, b half of c and d. BUt i'm getting abit stuck proving c again. Iill show the question then what i've got so far
A man standing at the top ad near the edge of a cliff of height 30m throws a ball into the air almost vertically upwars at a velocity of 12m/s. The ball then passes close to the man on he way down, hitting the ground at the foot of the cliff.
Calculate:
a) The maximum height above the cliff top
b) The time that elapses before the ball pases the man on the way down (choose another method of calculation and use it to check your answer)
c)the total time takedn for the ball to reach the ground. (as in part b choose another method to check your answer)
d) The velocity of the ball when it strikes the ground
answers
a) using v^2=u^2 +2as i rearranged for S and got 7.34m
b)using v=u+at i got 1.22s uptime and the same for the down time total 2.44s - proved it again using s=(u+V)t /2
c) used s=ut+1/2xat^2 , rearranged into a quadratic and got the values of 3.98s and -1.54s (usedthe 3.98 secs + 2.44 = 6.42s) - however i couldn't see another way of proving this without doing part d and then going back.
d) used v=u+at using u=0 a=9.81 t=5.21s and got 51.012m/s
I tried to go back using the final velocity to prove my answer in part c but all the equation give me different answers. can anyone spot where i've went wrong?
Please help, be looking for my mistake for about an hour now!
You are confusing speed with velocity.
Velocity is a vector - it has magnitude and direction.
In straight line motion the direction of velocity is specified by +ve or -ve. If you define up as the positive direction then -ve is the downwards direction. Hence -27 m/s means a speed of 27 m/s downwards.