Question

A person initially at rest throws a ball upward at an angle θ0= 75 ∘ with an initial speed v0=15 m/s . He tries to catch up to the ball by accelerating with a constant acceleration a for a time interval of 1.03 s and then continues to run at a constant speed for the rest of the trip. He catches the ball at exactly the same height he threw it. Let g= 9.81 m/s2 be the gravitational constant. What was the person's acceleration a (in m/s2)?

Answer 1

let's separate vertical and horizontal motion.
consider vertical. (we want to find something common in both motion to get a breakthrough, so that is time t.) we know \[u_{0, y} = 15\sin 75^{o}\] \[s =0\] \[a = -g\]
find t using \[s = u _{o} t + \frac{ 1 }{ 2 } at ^{2}\]

Answer 2

(sorry that u0 should be v0)
I will skip some step but you could find
\[t=\frac{ 2v _{0} \sin \theta }{ g }\]
then consider horizontal motion of the ball. you get the range R from t
\[R=\frac{ v _{0} ^{2} \sin 2\theta}{ g }\]

this is the same as the distance run by the man

Answer 3

now consider only the person. see what you have got from the above steps
then I will leave that to you :)
happy study :)

Answer 4

ref

sorry for bad drawing :(