Physics Question : A ball is thrown from the top of a building... / Speed of ball?

Question

A ball is thrown from the top of a building upward at an angle of 38◦ to the horizontal and with an initial speed of 17 m/s, as in the figure. The ball is thrown at a height of 46 m above the ground.
The acceleration of gravity is 9.8 m/s2 .

a)How long is the ball “in flight”? Answer in units of s.
b)What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

Any help would be amazing!

Answer 1

a)

The general equation giving the Y component of the displacement of a projectile fired with a velocity u and at an angle of θ with the horizontal ( vertical component u sin θ ) is ,

Y = ( u sin θ ) t - 0.5 g t^2

- 46 = ( 17 sin 38 ) t - 0.5 x 9.8 t^2

(the negative sign for Y shows that it is measured downwards from the starting point)

Or,

4.9 t^2 - 10.466 t - 46 = 0

Solving for t gives,

t = 4.313 s

This is the time for which the ball is in flight.



b)

Use the law of conservation of energy.



Energy of the projectile at the point of projection is,

0.5 m u^2 + mgh



The energy of the projectile at the time of hitting the ground is ,

0.5 m v^2 , where v is the final speed of the ball



Equating the two,

u^2 + 2 g h = v^2

17^2 + 2 x 9.8 x 46 = v^2

v = 34.5 m/s

Answer 2

A ball is thrown from the top of a building upward at an angle of 38◦ to the horizontal and with an initial speed of 17 m/s, as in the figure. The ball is thrown at a height of 46 m above the ground.

The acceleration of gravity is 9.8 m/s2 .



a)How long is the ball “in flight”? Answer in units of s.



17sin(38) = initial vertical velocity = 10.466 m/s



10.466^2 = 2gh



10.466^2/2g = h



109.542/19.6 = 5.59 m



time-up = 10.466/9.8 = 1.068 secs



Total height = 5.59 + 46 = 51.59 m



51.59/(1/2g) = t^2

51.59/4.9 = 10.528, sq-root = t = 3.245 secs (down-time)



Total time = 1.068 up & 3.245 down = 4.313 secs



b)What is the speed of the ball just before it strikes the ground? Answer in units of m/s.



3.245 x 9.8 = 31.8 m/s



17cos(38) = horizontal velocity = 13.4 m/s



sq-root(31.8^2 + 13.4^2) = 34.5 m/s (answer)

Answer 3

Use the equation h = ut + a million/2at^2 the place h is the top of the development = fifty one m, u = 0 is the preliminary downward velocity, t = time and a = 9.8 m/sec^2, the acceleration via gravity subsequently fifty one = 0 + 9.8/2 t^2 remedy for t: t = Sqrt(fifty one * 2 / 9.8) = 3.23 sec enable V = horizontal velocity. Then Vt = 35 m, the place t is the time of flight computed above So, V = 35/3.23 = 10.80 5 m/sec