Consider a rock thrown off a bridge of height 78.9 m at an angle = 25° with resp

Question

Consider a rock thrown off a bridge of height 78.9 m at an angle = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 10.2 m/s. Find the following quantities:

(a) the maximum height reached by the rock

(b) the time it takes the rock to reach its maximum height

(c) the time at which the rock lands

(d) how far horizontally from the bridge the rock lands

(e) the velocity of the rock (magnitude and direction) just before it lands.

magnitude

direction: Give your answer in degrees. If you think the answer is "20 degrees down from the positive x-axis", you would enter "-20" (note the negative sign!)

Explanation / Answer

As given in the question,

Height of bridge: s = 78.9 m

Angle with horizontal: = 25°

Initial speed of rock: u = 10.2 m/s

x-direction initial speed: u(x) = u cos = 10.2 cos25° = 9.24 m/s

y-direction initial speed:   u(y)  = u sin = 10.2 sin25° = 4.31 m/s

(a) For the maximum height reached by the rock,

Final y-direcionals peed at the topmost position = 0

using v^2 = u(y)^2 - 2*g*h

=> 0 = 4.31^2 - 2*9.8*h =>   h = 0.87 m

So, the total height required, H = s + h = 78.9 + 0.87 = 79.77 m

(b) For the time it takes the rock to reach its maximum height,

Using v = u(y) - g*t

=> 0 = 4.31 - 9.8*t =>   t = 0.439 s

(c) For the time at which the rock lands,

First we need to find the time it will take to reach from top position to the ground.

Using H = u(top)*T + (1/2)*g*T^2   , here u(top) = 0 at top position.

=> 79.77 = 0 + (1/2)*9.8*T^2 => T = 4.035 s

So, the total time required = t + T  = 0.439 + 4.035 = 4.474 s

(d) For the horizontal distance from the bridge where rock lands,

  d =  u(x)*(t + T) =  9.24*4.474 = 41.34 m

(e) For the velocity of the rock (magnitude and direction) just before it lands,

v(x) = u(x) = 9.24 m/s

v(y) = u(y) + g*T = 0 + 9.8*4.035 = 39.54 m/s

So, v = v(x) i^ + v(y) j^ =  9.24 i^ + 39.54 j^ m/s

Magnitude = sqrt[ v(x)^2 + v(y)^2 ] = 40.61 m/s

Since, the figure is not given so we can' find the exact angle for the final velocity.

It will be either - or 180° + depending on the figure.

where   = tan^-1 [ v(y) / v(x) ] = tan^-1 [ 39.54 / 9.24 ] = tan^-1 [4.279] = 76.85°

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

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