Question

A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in Figure 4-36, where t = 0 at the instant the ball is struck. The scaling on the vertical axis is set by va=16 m/s and vb=32 m/s. (a) How far does the golf ball travel horizontally before returning to ground level? (b) What is the maximum height above ground level attained by the ball?

Answer 1

somebody

Answer 2

till t=5 seconds u want to know in the question?

Answer 3

need working...need x...graph is confusing...

Answer 4

that is till 5 seconds right?

Answer 5

that straight line in the v-t graph represents.......?

Answer 6

ah yes according to the graph

Answer 7

it is 5 seconds

Answer 8

represents...................?above question answer with wat u know

Answer 9

v0=32-16=16m/s
t=5s
g=-9.8m/s^2
x-x0=v0t
y-y0=v0t-0.5gt^2

Answer 10

but i dunno if v0 is 16m/s...dunno if thats right

Answer 11

divide the sum into 2 parts first one till veocity becomes zero then when velocity starts from zero then add the displacements to get the total

Answer 12

sorry dont add them ......find the 2nd part alone that will give u the required answer

Answer 13

u get it?

Answer 14

nope...doesn't sounds right

Answer 15

u have the answer? we will work it out

Answer 16

nope...not for this question...

Answer 17

the entire thing is like this: the golf is at all times under the influence of gravity so its path wud be like this

so from that path the VERTICAL DISPLACEWMENT IS MAXIMUM AT t=2.5 seconds
now for the favourite part:
for horizontal range there is a formula: u^2sin2@/g(i can show the proof if u want)

for vertical displacemet after u find it,
use this formula: H=u^2*sin^2@/2g

now find the angle @(theta)]
]thisa is a projectile sum