Can some please help

do you have an equation list? if so look for the equation that models h(t) (height in reference to time) and fill it with your given values, one thing you'll need to work out as well is the max height(m), which will be using sin(11)=x/90

hope this helps

Could you use suvat ?? So:
S =
U = 68
V = /
A = 9.81
T = ?
So you said find the max height, to do this do I do..
Sin (11) = x/90 —> x = sin(11) x 90 = 17.17
S = 1/2 x (u + v) x t —-> 17.17 = 1/2 x (68 + 0) x t —-> 17.17 = 34 x t
But then I got 17.17/34 = 0.5 which isn’t 1.3s so I’m a bit confsued

Using two SUVAT equations to work out the vertical displacement. At the highest position, the velocity is equal to 0ms^-1. Find the maximum height using $V^2=u^2+2aS$, then sub into $S = u + at$ to show time is 1.255..., therefore 1.3s.

Sorry I still understand? @0le

Can you explain it a bit more please

How have you arrived at your second equation? I also think you need to take more care in this problem since you have horizontal and vertical components at play.


Image from here:
https://jameskennedymonash.wordpress.com/suvat-rearranged-3/



https://www.physicsclassroom.com/class/vectors/Lesson-2/Non-Horizontally-Launched-Projectiles-Problem-Solv

In fact, you don't even need to do calculus. You can effectively just state the vertical component of the velocity is zero and use one of the equations listed on that page website I listed. This is probably what they want you to do and both the calculus method and the substitution method give $\mathrm{t} = 1.322\mathrm{s}$.

@0le @ThiagoBrigido

So are you both saying to do this:

Horizontal component= 68 x cos(11) = 66.75064847
Vertical component = 68 x sin(11) = 12.97501169

Then @0le the website you have said use (V_fy = V_iy + a_y x t)

So (12.975...) = (66.75...) + (-9.81) x t

So -53.775... = -9.81 x t

So t = -5.48? But that’s wrong so idk what I’ve done wrong here

An archer fires an arrow towards a target as shown below
The diagram isn’t drawn to scale
The centre of the target is at the same height as the initial post ion of the arrow
The target is a distance of 90m from the arrow
The arrow has an initial velocity of 68ms-1 and is fired at an angle of 11 degrees to the horizontal

Air resistance has negligible effect on the motion of the arrow.

Show that the time taken for the arrow to reach its maximum height is about 1.3s.

I’m confused on how to do this at all, could someone please help

You don't need any suvats - just the s=d/t equation. Find the horizontal component of velocity by 68cos(11) and then do the following calculcation -> t=d/s = 90/68cos(11) = 1.348....s.
I hope that helps

@0le

So I know that Vfy = final velocity and Vfi = initial velocity

So is Vfy = 0 and Vfi = 68

So is it

68 = 0 x (-9.81) x t

So t = 6.93?

I don’t know what I’ve done