Projectile questions knock youself out.?

Question

1)(a) Prove that for a projectile fired from level ground at an angle θ above the horizontal, the ratio of the maximum height H to the range R is given by H/R = ¼ tan θ . (b) For what angle θ does H= R.

2)A movie stuntman is to run across and directly off a rooftop, to land on the roof of the next building 6.2 m away, 4.8m lower from the other building. Can he land there if he runs at 4.5m/s?

3) An archer tries to hit a target that is 20 m away from him. He can release the arrow at 25 m/s. Neglecting air resistance, estimate the angle at which the archer should aim to compensate for the fall of the arrow due to gravity

4)A golfer hits his tee shot along a flat fairway at 40o to the horizontal with an initial speed of 50 m/s. What is the range of the ball? What is the maximum height to which the ball rises?

5) A stone is thrown horizontally from the top of the building 440 m high. (a) Find its velocity after 3 seconds (b) How far is it from the ground at this time?

Answer 1

I don't think the first answerer should have given you all the answers, but rather just hints at how to do the problems; if you don't do them yourself you will never learn how.

Besides that, I have to point out that the answers can't possibly have more accuracy that the initial data, and since these data are only given to two figures all the answers should be rounded to two figures. The first guy is in love with his calculator, apparently.

Answer 2

1)

½R = VcosΘ x T ---->(1)

Where T is time taken to reach max height .

(VsinΘ)² = 2gH . --> (2)

H = -½gT² + VsinΘxT --->(3)

VsinΘ = gT ----> (4)

Substitute (4) in (2)

(gT)² = 2gH

gT² = 2H -----> (5)

Substitute (5) in (3)

H = -½(2H) + VsinΘ x T

2H = VsinΘ x T

TanΘ = SinΘ/CosΘ

sinΘ = 2H/VT

cosΘ = ½R/VT

TanΘ = (2H/VT)/(½R/VT) = 4H/R

H/R = ¼TanΘ .

b)

When H = R

¼TanΘ = 1

TanΘ = 4

Θ = 75.96°

2)

First lets calculate the time it will take him to fall a distance of 4.8m Vertically .

Vertical Motion :-

s = ½at² + ut

-4.8 = ½(-9.8)t² + 0

t = 0.9897433186 s

Min Speed Required = Distance / Time Taken

= 6.2 / 0.9897433186 = 6.264250421m

So no , he wont make it .

3)

It means that vertical displacement should be equal to zero .

s = ½at² + ut

0 = ½(-9.8)t² + 25sinΘxt

20 = 25cosΘ x t

t = 20/(25cosΘ)

0 = ½(-9.8)(20/25cosΘ)² + 25sinΘ(20/25cosΘ)

0 = -392/125 sec²Θ + 20TanΘ

sec²Θ = Tan²Θ + 1

0 = -392/125 Tan²Θ + 20TanΘ - 392/125

TanΘ = 6.216693814

TanΘ = 0.1608572064

Θ = 80.86184113 ( Not appropriate )

Θ = 9.138158873 ANS :)  

4)

Horizontal Component Velocity = 50cos40

Vertical Component Velocity = 50sin40

R is range and H is max Height

At max height V=0

0 = 50sin40 - 9.8T

T = 3.279528621 s

H = ½(-9.8)( 3.279528621)² + 50sin40(3.279528621)

H = 52.70285996 m

R = 50cos40 x 2T

R = 50cos40 x 2(3.279528621) = 251.2264676 m

5)

I think there is an info for speed of projection is missing or you meant thrown vertically instead of horizontally . Please Check .