A motorist speeds horizontally off a 50.0 meter high cliff. If the motorist want

Question

A motorist speeds horizontally off a 50.0 meter high cliff. If the motorist wants to land on level ground below, 90.0 m from the base of the cliff. (a) For how long (how many seconds) will the motorist be traveling in the air? (b) How fast must the motorcycle leave the cliff top to make this landing? (c) What is the final velocity (please indicate both magnitude and angle)?

This is the second time i post this, without a worked out part C. I need to see all the steps I especially need help with part C. please give magnitude and angle along with the steps I highly appreciate it thank you! :)

Explanation / Answer

Given,

Height = h = 50m ; horizontal distance = D = 90 m ;

(a)Let "t" be the time for which he will be in air.

we know from equation of motion that,

S = u t + 1/2 a t2

in our case, S = H = 50 m ; u = 0 and a = g = 9.8 m/s2, t is to be calculated.

1/2 g t2 = H => t = sqrt (2H/g)

t = sqrt (2 x 50 / 9.8) = 3.2 s.

Hence, t = 3.2 s

b)we know that,

speed = v = distance/ time = D/t

v = D/t = 90m/3.2 s = 28.13 m/s

hence required speed = v = 28.13 m/s.

c)Le V be the final velocity.

from equation of motion:

v2 = u2 + 2 a S

In our case, u = v = 28.13 m/s ; a = g = 9.8 m/s2 ; S = D = 90 m, v = V

V2 = 791.3 + 2 x 9.8 x 90 => V = 50.55 m/s

Hence, V = 50.55 m/s .

[Note: there is no angle given in the question so cant compute the X and Y component of this velocity and hence the angle].

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.