Please help! Projectile motion (1 Viewer)

[jacksonhughes95]

I really have no idea of how to do this at all haha, I'd really appreciate some help

 

[DamTameNaken]

In general you have to be familiar with your SUVAT equations and know that horizontal component is ucos(theta) and vertical component is usin(theta)

For Q1 a. We only have to consider the vertical component, it starts at a height of 100 m and accelerates downwards at 9.8m/s^2
Writing down our known values:
s= 100m
a= 9.8 m/s^2
vertical u= 0 m/s (Our vertical speed starts at 0m/s)
vertical v =?
t= ? (we're solving for t)

Now we can use any number of SUVAT equations to solve for t, I think it would be easiest to find 'v' and then solve for t using v= u + at

solving for v: v^2 = u^2 + 2as
v^2 = 0 + 2 x 9.8 x 100
V^2 = 1960
v = 44.27

sub in our 'v' value into v=u + at
44.27 = 0 + 9.8 x t
t= 4.52 seconds.

Now for part b. our values are:
t = 4.52 seconds
horizontal 'u' = 40m/s
horizontal 'v' = 40m/s (there's nothing to decelerate it
a = 0
s = ? (we're solving for s)

Now we can choose the most appropriate suvat equation

s = ut + (1/2)(at^2)
since a = 0, (1/2)(at^2) is equal to 0
s= ut
s= 40 x 4.52
s =180.8 m

 

[DamTameNaken]

For Q2. we have to split the vector into vertical and horizontal components.

Our vertical vector = sin30 x 145 = 72.5
Our horizontal vector = cos 30 x 145 = 125.57

for a. we only consider our vertical component, remember that when a ball reaches its maximum height its vertical velocity is equal to 0.
vertical v = 0
vertical u = 72.5
a = 9.8 m/s^2 downwards
t = ?
s = ? (we'r solving for s)

you can find 't' using v= u + at
and then you can solve for s using 's = ut + (1/2)(at^2)'

For b. you can cheat the question and don't have to do any working out by simply writing 'parabolas are symmetrical' and double your 't' value for Q2. a.

c. is just more SUVATS, use your 't' value from b. and the equation s= ut + (1/2)(at^2) where a is = 0.