Question

person kicks ball directed 53 degree above horizontal .ball lands on roof 7.2 m high.wall of building is 25 m away,and it takes ball 2.1s to pass directly over wall. calculate a)initial velocity of ball b)horizontal range of ball c)by what vertical distance does ball clear wall of building?

Answer 1

like that?

Answer 2

yup..

Answer 3

what's that vector (black, horizontal) in terms of the angle and V?

Answer 4

V is the red vector

Answer 5

\[v_{x}..\]

Answer 6

Vcos(53) is vx

25m = Vcos(53) * t

Answer 7

"it takes ball 2.1s to pass directly over wall"

25m = Vcos(53) * (2.1)

Answer 8

so V=?

Answer 9

vx=11.9m/s

Answer 10

check again...

Answer 11

divide both sides by 2.1 ...
divide both sides by cos(53)...

Answer 12

vx=vcos theta
11.9=v cos53
v=20 m/s

Answer 13

this is initial velocity of balll right??

Answer 14

19.8 ok close enough I guess...

Answer 15

so that's a)
b)...

Answer 16

now total horizontal range dx...???

Answer 17

ball starts at y=0 ends at y=7.2m (on the roof)
so:
7.2 = Vsin(53) *t - 1/2 * g * t^2

Answer 18

if you can find that t, you can find horizontal range... (because we already know horizontal velocity)

Answer 19

are you able to solve that for t? (hint: it's a quadratic equation)

Answer 20

m tryin..1 sec..

Answer 21

set it equal to zero
a = -4.9
b= 15.8
c= -7.2

Answer 22

approx.3.67 s=4s

Answer 23

2.67s

Answer 24

i took a=4.9,b=-16,c=-7.2

Answer 25

b^2 = 250
-4ac = -141.1
b^2 -4ac = 108.9
sqrt(b^2 -4ac) = 10.4
-b -sqrt(b^2 -4ac) = -15.8 -10.4 =-26.2
2a = - 9.8

-26.2/-9.8 = 2.67s

Answer 26

sign errors reposted.

Answer 27

so now you can find horizontal range

range = 11.9*2.67

Answer 28

your c should be+7.2 ryt??

Answer 29

-7.2

Answer 30

7.2 =Vsin(53) *t - 1/2 * g * t^2

Answer 31

set equal to zero
0 = -7.2 +Vsin(53) *t - 1/2 * g * t^2

Answer 32

ok now i got my mistake..sorry

Answer 33

cool.