A diver standing on a 50 feet high diving ...?

Question

A diver standing on a 50 feet high diving board calculates that he would be traveling at a speed of 87 miles per hour when he hits the water, with the acceleration of gravity being 32.2 feet/sec^2. Is the diver correct in his calculation? How fast in miles per hour would he be going when he hits the water (if he ever jumps)?

Answer 1

v^2 = 2gd in this case as the initial velocity is 0, and g is the acceleration due to gravity and d is the distance travelled.
ie. v^2 = 2* 32.2* 50 = 3220
Hence v = 56.75 ft/sec
To convert this to miles per hour, multiply this by 60*60 and divide by 5280
=38.69 mph

Answer 2

87 mph = 87 * (5280 feet/mile) / (3600 sec / hr) = 127.6 ft / sec

If would take 3.96 seconds to achieve that velocity

the distance you would cover is d = (1/2) a * t^2 = 252 ft, so unless the diver is one heck of a high-jumper he won't achieve that entry speed.

From a height of 50 feet, the correct time for entry is t=(2d/a)^2=1.762 sec. Velocity would be 56.74 ft/sec or 38.69 mph.

Answer 3

v^2=u^2+2as=2(32.2)(50)=3220
v=56.75 feet/s but i don't know how to convert to mile/hr.

Answer 4

without preliminary speed or assuming freefall, given: a=50ft g=32.2ft/s^2 Req'd: Vf=? fmla: a=a million/2gt^2 Vf=gt Sol'n: a=a million/2gt^2 subsequently t= sqrt of (2a/g) t=sqrt of (2*50ft/32.2ft/sec^2) t=a million.76sec Vf=gt =32.2ft/sec^2*a million.76sec =fifty six.672ft/sec* 3600sec/hr =204019.2ft/hr*1m/3.28ft*1km/1000m*1mile... Vf= 38.87mi/hr