If an object is propelled vertically upward with an initial velocity of \( v \) foet per second, the height h of the object after \( t \) seconds, at \[ h=v t=\frac{1}{2}(32) t^{2} \] Calculate the height h of the object when the initial velocity is 70 feet per secend and the time is 2 seconds The height of the object will be \( \square \) feet after 2 seconds

To calculate the height \( h \) of the object after 2 seconds with an initial velocity of 70 feet per second, we can use the formula for height: \[ h = v t - \frac{1}{2} (32) t^2 \] where: - \( v = 70 \) feet per second (initial velocity), - \( t = 2 \) seconds (time), - \( 32 \) is the acceleration due to gravity in feet per second squared. Now, substituting the values into the formula: \[ h = 70(2) - \frac{1}{2}(32)(2^2) \] Calculating each term: 1. Calculate \( 70(2) \): \[ 70 \times 2 = 140 \] 2. Calculate \( \frac{1}{2}(32)(2^2) \): - First, calculate \( 2^2 \): \[ 2^2 = 4 \] - Then, calculate \( 32 \times 4 \): \[ 32 \times 4 = 128 \] - Now, calculate \( \frac{1}{2}(128) \): \[ \frac{1}{2} \times 128 = 64 \] Now, substitute these values back into the height equation: \[ h = 140 - 64 \] Finally, calculate \( h \): \[ h = 76 \] Thus, the height of the object after 2 seconds will be \( \boxed{76} \) feet.