Question
A ball is thrown toward a cliff of height h with a speed of 28m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 3.3s later.
a) How high is the cliff? i got -29.2 m but my answer is wrong
b) What was the maximum height of the ball? im not sure how to solve this part
please help, thanks in advance
a) How high is the cliff? i got -29.2 m but my answer is wrong
b) What was the maximum height of the ball? im not sure how to solve this part
please help, thanks in advance
Answers
Vi = 28 sin 60 =24.25 m/s
v = Vi - 9.8 t
h = Vi t - 4.9 t^2
= 24.25(3.3) - 4.9(3.3)^2
= 26.7 meters high
not 29.2
when does v = 0 (top of arc)?
0 = 24.25 - 9.8 t
t = 2.47 seconds
so
h at top = 24.25(2.47) - 4.9 (2.47)^2
v = Vi - 9.8 t
h = Vi t - 4.9 t^2
= 24.25(3.3) - 4.9(3.3)^2
= 26.7 meters high
not 29.2
when does v = 0 (top of arc)?
0 = 24.25 - 9.8 t
t = 2.47 seconds
so
h at top = 24.25(2.47) - 4.9 (2.47)^2
Thanks Damon,
I realized that I forgot to multiply my velocity by time (3.3).
I tried to do the next part but my answer is not right. can you double check this?
c) What is the ball's impact speed?
I found fall time 3.3-2.47=0.83s
then I used this formula V1=V0+gt = 0+9.80*0.83 = 8.13m/s
where am I wrong?
Thanks
I realized that I forgot to multiply my velocity by time (3.3).
I tried to do the next part but my answer is not right. can you double check this?
c) What is the ball's impact speed?
I found fall time 3.3-2.47=0.83s
then I used this formula V1=V0+gt = 0+9.80*0.83 = 8.13m/s
where am I wrong?
Thanks