Question

Help me solve this please

Answer 1

Since it's a projectile being launched the only force acting upon it is gravity, since the object is in free fall once it's launched

so to calculate time you'd utilize the general formula of
$xf = xi + vxi(t) + \frac{1}{2} (a) {t}^{2}$
and then solve using time and make it into the y axis, so change the x's to y's, which will change a to g.
since Vyi is always usually 0, you can drop that out of this equation so the formula to find time would be
$t = \sqrt{ \frac{2(y)}{g} }$
So you'll plug in and it'll be
$t = \sqrt{ \frac{2( - 49m)}{ - 9.81 \frac{m}{s {}^{2} } } }$
to find the maximum height you'll have to do some trigonometry to solve it.
To make it easier draw a triangle
put the 60° mark as shown in the picture.
Then you'll need to find the hypotenuse or horizontal to find the vertical
So the hypotenuse would be the 113m/s
so then you'll use
$\ \sin( 60) = ( \frac{o}{h})$

plug in the numbers
$113( \ \sin (60) ) = o$
now that you have the vertical

use the formula
${vf}^{2} = {vi}^{2} + 2gd$
solve for d which will give you the hypotenuse
$d = { (-vertical \frac{m}{s} )}^{2} \div 2( - 9.81 \frac{m}{ {s}^{2} } )$
The "vertical" is what you found in the previous step.
Vf^2 is equal to 0 so you can just drop that number out since it's 0
then once you have that then youre not done yet
since you're on a cliff of 49 m you'll have to add 49m to the previous answer that you found d to find the maximum height.

I hope this helps!