A body is thrown vertically upwards with an initial velocity of 100m / s, after 4s of the launch its velocity is 60 m / s.

1. What is the maximum height reached?
2. In what time does the mobile travel that distance?
3. How long does it take to return to the starting point since it was launched?
4. How long does it take to reach heights of 300m and 600m?

Use g = 10m / s²

Question

A body is thrown vertically upwards with an initial velocity of 100m / s, after 4s of the launch its velocity is 60 m / s.

1. What is the maximum height reached?
2. In what time does the mobile travel that distance?
3. How long does it take to return to the starting point since it was launched?
4. How long does it take to reach heights of 300m and 600m?

Use g = 10m / s²

SmokeyBrown · Answer

You can model the height of this object with the equation:
$$Height = 100x - 10x ^{2} + 0$$
Because the initial velocity of the object is 100 m/s upwards, while gravity accelerates the object downward at a rate of 10 m/s^2; the initial height is not specified, so I assume it starts at ground level, 0.

Using this equation, can you answer the four questions? Please let me know if you need any more pointers

SmokeyBrown · Answer

Thank you for the corrections.
Indeed, the height would change according to 1/2*g*t^2, not g*t^2, as I mistakenly said.
Do not forget to include the initial velocity in the equation as well, since that will also affect the height. I think the full equation will look like
height = 100t - 1/2*(10*t^2), where t is the time in seconds
When I attempted to graph this equation, it seemed like the maximum height was 500, and the height never actually reached 600. This may have again been my error, so I'd be interested in seeing whether anyone will verify or refute that

Florisalreadytaken · Answer

i just got back to this -- i dont know what the Ice Cream i was doing last night haha -- most likely didnt read the question properly no, youre right, max height is 500 -- i got that as well 1$ v^2=u^2−2gh $ $$ 0^2=(100 m/s)^2−2(10 m/s^2)h $$ $$ h=500 $$

2) we know that $ v=u-gt $ $$ 0=100-10t $$ $$ t=10 $$

3) $ t_{total}= 2t $ $$ t_{total}=2 \times 10 $$ $$ t_{total}=20 $$

4) since everything is constant, we can just set a proportion for 300m we do: $$ \frac{500}{10}=\frac{300}{x} $$ $$ 500x=3000 $$ $$ x=6 \ sec $$ for 600 it would not be possible, since 500 is MAX.

Florisalreadytaken · Answer

wait actually 4 is a little deeper than it looks -- not everything is constant as we do have a change in velocity :\

SO final answer we can say that the time taken by the ball to reach at height h=300 can be found like this:
$$  
V^2=u^2-2 g h \
V^2=(100)^2-(2)(10)(300) \
V^2=10000-6000 \
V^2=4000 \
V=20 \sqrt{10} \ m/s $$

now, by the first equation of motion 
$$V=u-g t\
20 \sqrt{10}=100-(10) t\
10 t=100-20 \sqrt{10}\
t=10-2(\sqrt{10}) \
t=10-(2)(3.16) \
t=3.675 \mathrm{~s}$$
thats final answer 😹