Question

A ball is thrown upwards with an initial velocity of 33 m/s from a point 65◦ on the side of a hill which slopes upward uniformly at an angle of 28◦ . – At what distance up the slope does the ball strike? – Calculate the time of flight of the ball.​

Answer 1

Answer: 64.012 + 34.042 = 72.5 m

Step-by-step explanation:

S,(t) = (v cos 0)t

(1)

5, (t)= - =9t? + (usin )t + 5yo

Since initial launch height was 0, 5yo = 0

5,(t) = -

=g+? + (u sin0)t

(2)

h(t) = s›(t) tan ß

(3)

Pluggin (1) into (3)

h(t) = (vcos 0)(tan ß)t

Looking at the equations, we see that at the point of contact, sy(t) must equal h

Therefore,sy(t) = h(t)

- 59*? + (vsin 0)t = (u cos 0)(tanß)*

(4)

(-

29) +2 + (u sin 0)t - (v cos 0) (tan $) = 0

(

20

+ v[sin 0 - (cos 0) (tan B)]t = 0

Plugging in values:

1

(-2: 9.8) 12 + 331sin 65 - (cos 65)(tan 28)) = 0

-4.9 +2 + 22.49t = 0

Solving fort => t = 0 or t = 4.59 seconds

Only one solution: t = 4.59 seconds

Therefore, the range is:

Sr(5.8448) = (33 cos 65)(4.59) = 64.01 m

and the height is: h(5.8448) = S, tan 28 = 64.01 tan 28 = 34.04 m

and the distance up the slope is:

64.012 + 34.042 = 72.5 m