A stone is thrown vertically into the air at an initial velocity of \( 73 \mathrm{ft} / \mathrm{s} \). On a different planet, the height s (in feet) of the stone above the ground after t seconds is \( s=73 \mathrm{t}-5 \mathrm{t}^{2} \) and on Earth it is \( \mathrm{s}=73 \mathrm{t}-16 \mathrm{t}^{2} \). How much higher will the stone travel on the other planet than on Earth? When an object that is thrown upwards reaches its highest point (just before it starts to fall back to the ground), its velocity will be zero. Therefore, to find the maximum heights of the two objects, use the velocity equation, \( v=\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{f}^{\prime}(\mathrm{t}) \). The stone will travel \( \square \) (Do not round until the final answer. Then round to the nearest whole number as needed.) higher on the other planet.

To find the maximum height the stone reaches on both planets, we can set the velocity equations to zero and solve for time \( t \) on both planets. For the other planet, the height is given by: \[ s = 73t - 5t^2 \] Taking the derivative and setting it to zero: \[ v = \frac{ds}{dt} = 73 - 10t \] Setting \( v = 0 \): \[ 73 - 10t = 0 \implies t = 7.3 \text{ seconds} \] Substituting \( t = 7.3 \) back into the height equation: \[ s = 73(7.3) - 5(7.3)^2 = 532.9 - 266.45 = 266.45 \text{ feet} \] For Earth, the height is given by: \[ s = 73t - 16t^2 \] Taking the derivative: \[ v = \frac{ds}{dt} = 73 - 32t \] Setting \( v = 0 \): \[ 73 - 32t = 0 \implies t = 2.28125 \text{ seconds} \] Substituting \( t = 2.28125 \) back into the height equation: \[ s = 73(2.28125) - 16(2.28125)^2 = 166.75 - 83.125 = 83.625 \text{ feet} \] Now, subtract the maximum height on Earth from the maximum height on the other planet: \[ \text{Height difference} = 266.45 - 83.625 = 182.825 \text{ feet} \] Rounding this to the nearest whole number gives us that the stone will travel **183 feet** higher on the other planet than on Earth.