A hot air balloon is ascending vertically up ward at constant rate of 10 m/s. if

Question

A hot air balloon is ascending vertically up ward at constant rate of 10 m/s. if it was at 240 m above the ground when a key is dropped from the balloon at time t = 0 s, find the maximum height reached by the key from the ground, find the time 1 at which the key hits the ground, find the position of the key relative to the ground 5.0 seconds after being released from the balloon, in unit vector notation, find the velocity of the key relative to the balloon 5.0 seconds after being released from the balloon.

Explanation / Answer

a) key goes slightly upward and falls down. at the top point of key velocity of key will be 0. so u=10, v=0, a=-9.8 Hence v^2-u^2=2*a*S or S= 5.1 m.. Hence maximum height reached by key from ground will be 240+5.1 = 245.1 m ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; b) time for ascent of key + time for descent of key. for ascent, v=u+at; 0=10 -9.8t or t=1.02 sec. for descent: S=ut + 1/2at^2 or -245.1= 0 - 1/2*9.8*t^2 so t=7.07 sec. hence total time = 1.02+7.07 = 8.09 seconds.. ;;;;;;;;;;;;;;;;;; c) at 5 seconds key will be in descent. hence t = 5-1.02 = 3.98; S=ut + 1/2at^2 = 0 -1/2*9.8*3.98^2 = -77.6m.. Hence position of key relative to ground will be 245.1-77.6 = 167.48m ;;;;;;;;;;;;;;;;; d) v = u + at ; v = 0 - 9.8*3.98 = - 39 this will be v= -39 j m/s in unit vector notation.. let me know if anything is unclear

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