If i have an angle, distance to a barrier that is a certain height, how do i find the initial velocity/time?

Question

Given:
Angle that the projectile left the ground
Height of the wall
Distance to the wall

How do i find the initial velocity, the velocity of the projectile when it reaches the wall and the time until it reaches the wall?

Any help is greatly appreciated 'cause physics is killing me.

For example:
A home run is hit in such a way that the baseball just clears a wall 21 m high, located 130 m from home plate. The ball is hit at an angle of 35° to the horizontal, and air resistance is negligible. Find
(a) the initial speed of the ball,
m/s
(b) the time it takes the ball to reach the wall, and
s
(c) the velocity components and
m/s (x component),
m/s (y component)
the speed of the ball when it reaches the wall.
m/s

Answer 1

First: Recognize that velocity at an angle can be split into horizontal and vertical velocity. Same with distance.

d = vi + 1/2 at^2

1.) Use our vertical distance (in your supplied problem: 21m)
21 m = viv * t + 1/2 at^2
21 m = viv * t + -4.9t^2

2.) Use our horizontal distance (in your supplied problem: 130m)
130 m = vih * t + 1/2 at^2
130 m = vih * t since a = 0
t = 130m / vih

3.) Given the angle, we can use the tangent to find the ratio of vertical to horizontal velocity or distance (35):
tan a = opposite / adjacent = initial vertical / initial horizontal
tan 35 = .700 = viv / vih
vih = viv / .700 = 10/7 * viv

4.) Go back to step 1 and use the equation for we derived for the vertical distance. We'll convert the time equation we derived in step 2, and use the conversion from vih to viv we derived in step 3, so that we're down to 1 variable: viv:

21 m = viv * t + -4.9t^2
t = 130m / vih
t = 130m * 7/10 / viv = 91 / viv
21 m = viv * 91m / viv + -4.9(91m/viv)^2
21 m = 91 - 40576.9 / viv^2
70 = 40576.9 / viv^2
viv^2 = 40576.9 /70 = 579.67
viv = 24.0763 m/s

sin a = opposite / hypotenuse = vertical / total
sin a = .574 = vertical / total
total = vertical / .574
total = 24.0763 / .574 = 41.945 m/s (solution for a)

b.) t = 130m / vih
cos a = adj / hyp = horizontal / total
.819 = vih / vt
vih = .819 vt = 34.259 m/s
t = 130m / 34.259 m/s
t = 3.795 sec (solution for b)

c.) When the ball reaches the wall:
vv = viv + at = 24.0763 m/s + -9.8t = 24.0763 m/s + -9.8* 3.795 sec = 24.0763 m/s - 37.191 m/s = -10.115 m/s (going down)
vh = vih + at = vh (since a = 0) = 34.259 m/s

Answer 2

it is very difficult to answer this question with out a specific example to work

you cannot in fact, calculate the initial velocity, (even with the angle) unless you know how high up the wall the projectile hits

most students have trouble getting the notion of the two componants (up and over, x and y) of the ballistic motion

for most calculations, you have to split the two motions up and treat them separately

for example, the upward initial velocity is equal to the total initial velocity times the sin of the angle

the upward velocity (which depends on both angle and total velocity) determines how long the ballistic will be air borne

the sideways velocity doesn't matter for how long the projectile is up, it will just affect how far it goes during that time

this site has the primary equations that are used

http://zeta.lerc.nasa.gov/education/rocket/ballflght.html

good luck

again, if you were to provide an actual problem

with an angle and an intitial velocity
it would not be hard to show you where the projectile lands

again, remember that the up and down can be treated separately from the over

the up velocity will slowdown at decelleration "g" (32.2 ft/sec/sec)

the time to zero velocity is calcuable

you can work it backwords with the horizontal componant if the distance is provided