Urgent Help! What do I do???

How would I answer this question?

A cricket batsman hits a ball at a speed of 27m/s at an angle of 60 degrees to the horizontal. How far would you have to stand in order to catch it (assuming you have to catch it before it hits the ground)?

How would I attempt this one since I don't know time or dist???

Reply 1

Sinnoh

Split into horizontal and vertical vectors using trigonometric ratios. You know that the overall vertical displacement will be 0, the vertical acceleration -9.8ms^-2 and the initial vertical velocity is whatever the vertical vector is. With these three variables you can calculate the time taken and therefore use the horizontal velocity to calculate horizontal distance travelled, assuming constant horizontal velocity.

Reply 2

Yatayyat

What suvat equation do you use to calculate the time?

Reply 3

Yatayyat

I used the suvat equation of s = ut + 1/2 at^2. But not sure now since times comes out to be a negative value while the other is zero... 😕

If I did just use the time value and ignored the the minus sign it turns out to be a distance of 64m to 2 sig fig. Is this the right answer? I’m really not sure...

Reply 4

Sinnoh

Original post by Yatayyat

What suvat equation do you use to calculate the time?

Ok I have another idea. Use final velocity = 0 to calculate the time taken for it to reach its highest point and then calculate its drop separately

Reply 5

Yatayyat

Original post by Sinnoh

Ok I have another idea. Use final velocity = 0 to calculate the time taken for it to reach its highest point and then calculate its drop separately

How could I calculate the time taken to reach maximum displacement if I don't know the maximum height that is reached in the first place? I can't use 's = 1/2 ( u+v) * t' because I don't know the value of s, or neither can I use it for 's = ut + 1/2 at^2' since it doesn't even use the final velocity and again I don't know the value of S.

Reply 6

amaranApoc

You can calculate t by doing this
a = g = -9.8
u = 27sin(60)
v = -27sin(60) given that the velocity as it reaches the height it started at will be equal but opposite
then t = (v-u)/a

(edited 6 years ago)

Reply 7

Yatayyat

Original post by amaranApoc

You can calculate t by doing this
a = g = -9.8
u = 27sin(60)
v = -27sin(60) given that the velocity as it reaches the height it started at will be equal but opposite
then t = (v-u)/a

Okay so you are saying that to calculate time taken to reach max height (which means v=0 at that point), you do t = (0- (27sin(60)) / 9.8 which gives -2.38 seconds, doubling that time to gives the other half of the trajectory ( since you are assuming that it is equal and opposite) then gives a time of -4.77 seconds. It still gives time to be negative...

Reply 8

amaranApoc

Original post by Yatayyat

I'm suggesting you calculate the length of time for the entire arc, which is why i had V being equal to -27sin(60) but what you're doing would also work since it's the exact same thing.

The reason you're getting a negative value is because you've forgotten a negative in your acceleration. It's accelerating downwards so it's -9.8 rather than 9.8

Reply 9

Yatayyat

Original post by amaranApoc

Oh, I get what you mean now, so taking the entire journey to work out t with also taking g to be -9.8 gives 4.77 seconds.

It makes sense since the ball is first decelerating as it launches upwards therefore g must be -9.8 and a v = -27 sin(60) (as it has direction) and this gives t to be 2.38s. Doing a similar thing for the other half where the ball accelerate as it goes down is must have a g = +9.8 and v = 27 sin(60). Adding both times gives +4.77 s.

I'm guessing now to work out the horizontal distance that it travels has to be 4.77*27cos(60) (using s=vt) which gives an answer of 64m to 2 s.f