Physics questions- need answers + explanations?

Question

A projectile is fired with an initial speed of 30 m/s at an angle of 60 degrees above the horizontal. The object hits the ground 7.5 s later.
-to what maximum height above the launch point does the projectile rise?
-what are the magnitude and direction of the projectile's velocity at the instant it hits the ground?
ur playing right field for the baseball team.ur team is up by one run in the bottom of the last inning of the game when a ground ball slips through the infield and comes straight toward you. As u pick up the ball 65m from home plate, you see a runner rounding third base and heading for home with tying run.u throw the ball at an angle of 30degre. above the horizonal with just the right speed so that the ball is caught by the catcher, standing on home plate, at the same height as u threw it. As u release the ball, the runner is 20m from home plate and running full speed at 8m/s.Will the ball arrive in time for your team's catcher to make the tag and win the game?-give times

Answer 1

NICHOLAS L has good ideas, but let's just think a bit.
Whatever goes up must come down :)
It took time to go up and equal time to come down.
(A)
H=(1/2) gt^2
H - height
g- gravitational acceleration (9.81 m/s^2)
t- time (one way!) in your case it is 1/2 of 7.5 sec)
H=(1/2) 9.81 (7.5/2)=18.4 meters

(B)
When the objects hits the ground it will have both vertical and horizontal components. If there was no air resistance and the gun was on the ground level then the velocity the projectile hits the ground will be the of the same magnitude and a mirror reflection of the velocity it left the barrel (upward to left becomes downward to right)
So it is still 30m/s but the angle is 120 degrees (which is a mirror reflection of 60)

And now you are ready to answer the baseball question.

Answer 2

You have 3 formulas u need to know for these sort of questions.
s= ut + 1/2 at^2
v^2 = u^2 + 2as
a = (v-u)/ t
where
a is acceleration dun to free fall
v is the final speed
u is the initial speed
t is time taken for the projectile to hit the ground.
s is the max height in the vert dir or the max length in the horizontal
assuming no air resistence,
taking motion in 2dimensions,

vertical motion
a= 9.81 ms^-1
u= 30sin60
s= ut + 1/2 at^2
You can solve for s.
to find the velocity, you need to solve for v in both the vert and horizontal directions. Its pretty long so i wont do it. You need to use any one of the formulas to solve for v horizontal and v vertical. use pythagoras theorem to find the magnitude

Answer 3

Wherever you observed the concern, again up a couple of pages to discover the relevant equations. First, discover the second of inertia and angular momentum equations for a uniform disc (no longer "disk" -- that is a pc reminiscence gadget), and use it to discover the proper figures for the merry-pass-circular on my own. Now, while the child is walking alongside the tangent, his mositon is linear, so he does no longer give a contribution to angular momentum, nor to second of inertia. When he jumps on, he is a factor mass of 30kg at a radius of 2m. Note that the e-book made it less complicated for you, in that the children's linear pace fits the merry-pass-circular's angular pace. The pace is not going to difference while the child hops on: you are simply exchanging the linear momentum to angular, and the 2 are given as matching speeds on the primary factor. Computing his linear momentum is also less complicated for you (and you may also discover a few support in surroundings the 2 expressions same to 1 a different). Finding the rotational time is a practical subject of dividing the velocity (5m/sec) into the circumference (two*pi*2m).