← Back to learning objective

Time of impact

An object is dropped and its height above the ground (in meters) at time tt seconds is given by s(t)=100−4.9t2s(t) = 100 - 4.9t^2, until it strikes the ground.

Find the time tt at which the object hits the ground (that is, when s(t)=0s(t) = 0). Give the value of tt in seconds.

Hint

Set s(t)=0s(t) = 0 and solve for tt, keeping only the positive solution since time cannot be negative.

Step-by-step solution

Set s(t)=100−4.9t2=0s(t) = 100 - 4.9t^2 = 0. Solving gives t2=1004.9t^2 = \dfrac{100}{4.9}, so t=1004.9t = \sqrt{\dfrac{100}{4.9}}.

Since tt represents time, we discard the negative root, giving t≈4.5175t \approx 4.5175 seconds as the moment of impact.

Velocity at impact

Find the velocity of the object at the instant it strikes the ground.

Hint

Differentiate s(t)s(t) to get v(t)v(t), then substitute the impact time found earlier.

Step-by-step solution

Differentiate s(t)=100−4.9t2s(t) = 100 - 4.9t^2 to get v(t)=s′(t)=−9.8tv(t) = s'(t) = -9.8t.

Substituting t≈4.5175t \approx 4.5175 seconds gives v≈−9.8(4.5175)≈−44.27v \approx -9.8(4.5175) \approx -44.27 m/s, meaning the object is moving downward at about 44.27 m/s at impact.

Acceleration at impact

Find the acceleration of the object at the instant it strikes the ground.

Hint

Differentiate v(t)v(t) to get a(t)a(t); note that acceleration due to gravity here is constant.

Step-by-step solution

Differentiate v(t)=−9.8tv(t) = -9.8t to get a(t)=v′(t)=−9.8a(t) = v'(t) = -9.8 m/s2^2.

Since this is constant for all tt, the acceleration at the moment of impact is simply a=−9.8a = -9.8 m/s2^2, representing the constant downward pull of gravity.

Related problems

Practice similar Calculus problems interactively.

Start Closing the Gap