Question

A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant it is 64 feet above the ground.a. How many seconds after its release will the bag strike the ground?b. At what velocity will it hit the ground?

Answer 1

Answer:

a. 2.56 sec

b. 65.96 feet.

Step-by-step explanation:

a) The equation of motion of a body under gravity gives the equation  

$h = ut + gt^{2}$ ......... (1) {Where u is the initial velocity of the body and t is the time to descend h height and g is the acceleration due to gravity.

Now, from the equation (1) we have  

$64 = - 16t + 16 t^{2}$ {Since g = 32 feet/sec² }

⇒ $t^{2} - t - 4 = 0$

So, applying Sridhar Acharya formula  

$t = \frac{1 + \sqrt{ 1 + 16} }{2} = 2.56$ seconds {Neglecting the negative root as t can not be negative.} (Answer)

b) Again we have the formula that  

$v^{2} = u^{2} + 2gh$ {Where v is the final velocity before hitting the ground.}

⇒ $v^{2} = (-16)^{2} +2\times 32 \times 64 = 4352$

⇒ v = 65.96 feet / sec. (Answer)